Evaluate the following:

Example 01:

sin²0^c + sin²(π/6)^c + sin²(π/3)^c + sin²(π/2)^c  

= (0)² + (1/2)² + (√3/2)² + (1)²

= 0 + 1/4 + 3/4 + 1 = 1+ 1 = 2

Ans: sin²0^c + sin²(π/6)^c + sin²(π/3)^c + sin²(π/2)^c = 2

Example 02:

cos²0 + cos²(π/6)^c + cos²(π/3)^c + cos²(π/2)^c  

= (1)² + (√3/2)² + (1/2)² + (0)²

= 1 + 3/4 + 1/4 + 0 = 1+ 1 = 2

Ans: cos²0 + cos²(π/6)^c + cos²(π/3)^c + cos²(π/2)^c = 2

Example 03:

sin (π)^c + 2 cos (π)^c + 3 sin (3π/2)^c + 4 cos (3π/2)^c – 5 sec (π)^c + 6 cosec (π/2)^c

=  (0) + 2(-1) + 3(-1) + 4 (0) – 5 (-1) + 6 (1)

= 0 – 2 – 3 + 5 + 6 =  6

Ans: sin (π)^c + 2 cos (π)^c + 3 sin (3π/2)^c + 4 cos (3π/2)^c – 5 sec (π)^c + 6 cosec (π/2)^c = 6

Example 04:

sin (0)^c + 2 cos (0)^c + 3 sin (π/2)^c + 4 cos (π/2)^c + 5 sec (0)^c + 6 cosec (π/2)^c

= 0 + 2 (1) + 3 (1) + 4 (0) + 5 (1) + 6 (1)

= 2 + 3 + 0 + 5 + 6 = 16

Ans: sin (0)^c + 2 cos (0)^c + 3 sin (π/2)^c + 4 cos (π/2)^c + 5 sec (0)^c + 6 cosec (π/2)^c = 16

Example 05:

4 cot 45° – sec² 60° + sin² 30°

= 4 (1) – (2)² + (1/2)² 

=  4  – 4 + 1/4  = 1/4

Ans: 4 cot 45° – sec² 60° + sin² 30° = 1/4

Verify the Following:

Example 06:

• cot² 60° + sin² 45° + sin² 30° + cos² 90° = 13/12
• Solution:

L.H.S. = cot² 60° + sin² 45° + sin² 30° + cos² 90°

∴ L.H.S. = (1/√3)²  + (1/√2)² + (1/2)² + (0)²

∴ L.H.S. =  1/3 + 1/2 + 1/4 = 5/6 + 1/4

∴ L.H.S. = 26/24 = 13/12 = R.H.S.

∴   cot² 60° + sin² 45° + sin² 30° + cos² 90° = 13/12  (Proved)

Example 07:

• sin² 30° + cos² 60° + tan² 45° + sec² 60°  – cosec² 30° = 3/2
• Solution:

L.H.S. = sin² 30° + cos² 60° + tan² 45° + sec² 60°  – cosec² 30°

∴ L.H.S. = (1/2)²  + (1/2)² + (1)²  + (2)² – (2)²

∴ L.H.S. =  1/4 + 1/4 + 1 + 4 – 4 = 2/4 +1

∴ L.H.S. = 1/2 + 1 = 3/2  = R.H.S.

∴  sin² 30° + cos² 60° + tan² 45° + sec² 60°  – cosec² 30° = 3/2   (Proved)

Example 08:

• 4 cot² 30° + 9 sin² 60° – 6 cosec² 60°  -(9/4) tan² 60° = 4
• Solution:

L.H.S. = 4 cot² 30° + 9 sin² 60° – 6 cosec² 60°  – (9/4) tan² 60°

∴ L.H.S. = 4 (√3)²  + 9 (√3/2)² – 6 (2/√3)² – (9/4)(√3) ²

∴ L.H.S. = 4 x 3  + 9 (3/4) – 6 (4/3) – (9/4)(3)

∴ L.H.S. = 12  + 27/4 – 8 – 27/4 = 4 = R.H.S,

∴ 4 cot² 30° + 9 sin² 60° – 6 cosec² 60°  -(9/4) tan² 60° = 4 (Proved)

Example 09:

Example 10:

Example 11:

• If 2 cos²θ + 3cosθ = 2, then find cosθ.
• Solution:

Given 2 cos²θ + 3cosθ = 2  i.e 2 cos²θ + 3cosθ – 2 = 0

let cosθ = x

∴ 2 x² + 3x – 2 = 0

∴ 2 x² + 4x – x – 2 = 0

∴ 2 x(x + 2) – 1(x + 2) = 0

∴ (x + 2)(2x – 1) = 0

∴ (x + 2) = 0 or (2x – 1) = 0

∴ x = – 2 or x = 1/2

∴ cosθ = – 2 or cosθ = 1/2

Now – 1 ≤ cos θ ≤ 1, thus cosθ = – 2 is not possible.

∴ cosθ = 1/2\

Example 12:

• If 6sin²θ  – 11sinθ + 4 = 0, find cosθ.
• Solution:

Given 6sin²θ  – 11sinθ + 4 = 0

let sinθ = x

∴ 6x²  – 11x + 4 = 0

∴ 6x²  – 8x – 3x + 4 = 0

∴ 2x(3x  – 4) – 1(3x – 4) = 0

∴ (3x  – 4)(2x – 1) = 0

∴ 3x  – 4 = 0 or 2x – 1 = 0

∴ x = 4/3 or x = 1/2

Now – 1 ≤ sin θ ≤ 1, thus sinθ = 4/3 is not possible.

∴ sinθ = 1/2

Example 13:

• If 3tan²θ  – 4√3 tanθ + 3 = 0, then find tanθ.
• Solution:

Given 3tan²θ  – 4√3 tanθ + 3 = 0

let tanθ = x

∴ 3x²  – 4√3 x + 3 = 0

∴ 3x²  – 3√3 x –√3 x + 3 = 0

∴ √3 x √3 x²  – 3√3 x –√3 x + 3= 0

∴ √3 x ( √3x – 3 )  – 1(√3 x – 3) = 0

∴  (√3x – 3 )( √3 x – 1) = 0

 ∴ √3x – 3  = 0 or √3 x – 1 = 0

∴ x = 3/√3 or x = 1/√3

∴ x = √3 or x = 1/√3

∴  tanθ = √3 or tanθ = 1/√3

Example 14:

• If 4sin²θ  – 2(√3 + 1)sinθ + √3 = 0, then find sinθ. Hence find the angle θ.
• Solution:

Given 4sin²θ  – 2(√3 + 1)sinθ + √3 = 0

let sinθ = x

∴ 4x²  – 2(√3 + 1)x + √3 = 0

∴ 4x²  – 2√3x – 2x + √3 = 0

∴ 2x(2x  – √3) – 1(2x – √3 ) = 0

∴ (2x  – √3)(2x – 1) = 0

∴ 2x  – √3  = 0 or 2x – 1 = 0

∴ x = √3/2 or x = 1/2

∴  sinθ = √3/2 or sinθ = 1/2

∴ θ = sin^-1(√3/2) or θ = sin^-1(1/2)

∴ θ = (π/3)^c i.e. 60° or θ = (π/6)^c i.e. 30°

Example 15:

• Find the acute angles A and B satisfying cot (A + B) = 1 and cosec (A- B) = 2
• Solution:

Given cot (A + B) = 1 and cosec (A- B) = 2

we know that cot 45° = 1 and cosec 30° = 2

∴ A + B = 45° and A- B = 30°

Adding the two equations we get

2A = 75° i.e. A = 37.5°

Subtracting the two equations we get

2B = 15° i.e. B = 7.5°

∴ A = 37.5° and B = 7.5°

Example 16:

• Find the acute angles secA.cotB – secA – 2 cotB + 2 = 0
• Solution:

Given secA.cotB – secA – 2 cotB + 2 = 0

secA.(cotB – 1) – 2 (cotB – 1) = 0

(cotB – 1)(secA – 2) = 0

∴ cotB – 1 = 0  and/or secA – 2 = 0

∴ cotB = 1 and/or secA = 2

∴B =  cot^-11 and/or A = sec^-12

∴B = (π/4)^c i.e. 45° and/or A = (π/3)^c i.e. 60°

∴ A = (π/3)^c i.e. 60° and B = (π/4)^c i.e. 45°

The post Use of Trigonometric Functions Values of Standard Angles appeared first on The Fact Factor.

Signs of Trigonometric Ratios in Different Quadrants:

Diagram Showing Signs of Trigonometric Ratios in Different Quadrants

Diagram showing value of angle in different quadrants:

Examples Based on Signs of Trigonometric Ratios:

Example – 01:

State the signs of following Trigonometric Ratios (Functions)

sin 675^o

sin 675 ^o = sin (360^o + 315^o) = sin (315^o) = sin (270^o + 45^o)

Thus the angle 675^o lies in the fourth quadrant, where sin function is negative.

Hence sin 675 ^o is negative

sin 159^o

sin 159 ^o = sin (90^o + 69^o)

Thus the angle 159^o lies in second quadrant, where sin function is positive.

Hence sin 159 ^o is positive

cos 573 ^o

cos 573 ^o = cos (360^o + 213^o) = cos (213^o) = sin (180^o + 33^o)

Thus the angle 573^o lies in the third quadrant, where cos function is negative.

Hence cos 573 ^o is negative

cos 250 ^o

cos 250 ^o = cos (180^o + 70^o)

Thus the angle 250^o lies in the third quadrant, where cos function is negative.

Hence cos 250 ^o is negative

cos (-8π/3)^c

cos(-8π/3)^c = cos (-2π – 2π/3) = cos (- 2π/3) =  cos (- π/2 – π/6)

Thus the angle (-8π/3)^c lies in the third quadrant, where cos function is negative.

Hence cos(-8π/3)^c is negative

cos (11π/9)^c

cos(11π/9)^c = cos (π + 2π/9)

Thus the angle (11π/9)^c lies in the third quadrant, where cos function is negative.

Hence cos(11π/9)^c is negative

sec (3π/5)^c

sec(3π/5)^c = sec (π/2 + π/10)

Thus the angle (3π/5)^c lies in the second quadrant, where sec function is negative.

Hence sec(3π/5)^c is negative

Example – 02:

Determine the quadrant in which θ lies from following trigonometric functions.

sin θ > 0 and sec θ < 0

Given, sin θ > 0 and sec θ < 0. i.e. sin θ is positive and sec θ is negative

Hence θ lies in the second quadrant.

cos θ < 0 and cot θ > 0

Given, cos θ < 0 and cot θ > 0, i.e. cos θ is negative and cot θ is positive

Hence θ lies in the third quadrant.

sec θ > 0 and cosec θ < 0

Given, sec θ > 0 and cosec θ < 0, i.e. sec θ is positive and cosec θ is negative

Hence θ lies in the fourth quadrant.

Example – 04:

If tan θ= -2 and θ lies in second quadrant, find values of other trigonometric functions.

Solution:

Given tan θ = -2

We have, 1+ tan²θ = sec²θ

∴  1+ (-2)² = sec²θ

∴  5 = sec²θ

∴  sec θ = ± √5

Now θ lies in second quadrant where sec ratio is negative.

∴  sec θ = – √5

∴  cos θ = – 1/√5

Now, tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (-2) x ( – 1/√5)  = 2/√5

Now, cot  θ = 1/ tan θ = 1/-2 = – 1/2

cosec θ = 1/sin θ = 1/(2/√5) = √5/2

Example – 05:

If tan θ = 5/12 and π < θ < 3π/2, find values of other trigonometric functions.

Solution:

Given tan θ = 5/12 and π < θ < 3π/2

We have, 1+ tan²θ = sec²θ

∴  1+ (5/12)² = sec²θ

∴  1 + 25/144 = sec²θ

∴  (144 + 25)/144 = sec²θ

∴ 169/144 = sec²θ

∴  sec θ = ± 13/12

Now π < θ < 3π/2, hence θ lies in the third quadrant where sec ratio is negative.

∴  sec θ = – 13/12

∴  cos θ =1/(- 13/12) = – 12/13

Now, tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (5/12) x ( – 12/13)  =  – 5/13

Now, cot  θ = 1/ tan θ = 1/( 5/12 )= 12/5

cosec θ = 1/sin θ = 1/(- 5/13) = – 13/5

Example – 06:

If cos θ= 4/5 and 3π/2 < θ < 2π, find values of other trigonometric functions.

Solution:

Given cos θ= 4/5 and 3π/2 < θ < 2π

We have, sin²θ + cos²θ = 1

∴  sin²θ + (4/5)² = 1

∴  sin²θ = 1 – 16/25

∴  sin²θ = (25 – 16)/25

∴  sin²θ = 9/25

∴  sin θ = ± 3/5

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where sin ratio is negative.

∴  sin θ = – 3/5

Now, tan θ = sin θ / cos θ

∴  tan θ = (- 3/5) / (4/5) = – 3/4

Now, cosec θ = 1/sin θ = 1/(- 3/5) = – 5/3

sec θ = 1/cos θ = 1/(4/5) = 5/4

cot  θ = 1/ tan θ = 1/( -3/4 )= – 4/3

Example – 07:

If tan θ=  – 4/3 and 3π/2 < θ < 2π, find values of 3 sec θ + 5 tan θ.

Solution:

Given tan θ = – 4/3 and 3π/2 < θ < 2π

We have, 1+ tan²θ = sec²θ

∴  1+ (-4/3)² = sec²θ

∴  1 + 16/9 = sec²θ

∴  (9 + 16)/9 = sec²θ

∴ 25/9 = sec²θ

∴  sec θ = ± 5/3

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where sec ratio is positive.

∴  sec θ = 5/3

Now, 3 sec θ + 5 tan θ = 3 x (5/3) + 5 x (-4/3) = 5 – 20/3 = (15 – 20)/3 = – 5/3

Example – 08:

If sec θ= √2 and 3π/2 < θ < 2π, find values of

Solution:

Given sec θ= √2 and 3π/2 < θ < 2π

We have,1 + tan²θ = sec²θ

∴  1 + tan²θ = (√2)²

∴  1 + tan²θ = 2

∴  tan²θ = 1

∴  tan θ = ± 1

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where tan ratio is negative.

∴  tan θ = – 1

Now cot θ = 1/tan θ = 1 /(-1) = -1

cos θ = 1/sec θ = 1/√2

tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (-1) x ( 1/√2)  =  – 1/√2

cosec θ = 1/sin θ = 1/(- 1/√2) = – √2

Example – 09:

If cos θ= – 3/5 and π < θ < 3π/2, find value of

Solution:

Given cos θ= – 3/5 and π < θ < 3π/2

We have, sin²θ + cos²θ = 1

∴  sin²θ + (- 3/5)² = 1

∴  sin²θ = 1 – 9/25

∴  sin²θ = (25 – 9)/25

∴  sin²θ = 16/25

∴  sin θ = ± 4/5

Now π/ < θ < 3π/2, hence θ lies in the thir quadrant where sin ratio is negative.

∴  sin θ = – 4/5

Now, tan θ = sin θ / cos θ

∴  tan θ = (- 4/5) / (- 3/5) = 4/3

Now, cosec θ = 1/sin θ = 1/(- 4/5) = – 5/4

sec θ = 1/cos θ = 1/(- 3/5) = – 5/3

cot  θ = 1/ tan θ = 1/( 4/3 )= 3/4

Example – 10:

If  5 tan A = √7  where π < A  < 3π/2 and sec B =  √11  where 3π/2 < B  < 2π, find value of cosec A – tan B.

Solution:

Given 5 tan A = √7  where π < A  < 3π/2

∴ tan A = √7/5

∴ cot A = 5/√7

We have,1 + cot²A = cosec²A

∴  1 + (5/√7)² = cosec²A

∴  1 + (25/7) = cosec²A

∴  (7 + 25)/7 = cosec²A

∴  cosec²A = 32/7

∴ cosec A = ±  √32/√ 7 = ± 4√2/√ 7

Now π < A < 3π/2, hence A lies in the third quadrant where cosec ratio is negative.

cosec A = – 4√2/√ 7

Given sec B =  √11  where 3π/2 < B  < 2π

We have,1 + tan²B = sec²B

∴  1 + tan²B = (√11)²

∴  1 + tan²B = 11

∴  tan²B = 10

∴  tan B = ± √10

Now 3π/2 < B < 2π, hence B lies in the fourth quadrant where tan ratio is negative.

∴  tan B = – √10

The value of the quantity cosec A – tan B =  – 4√2/√7 – √10

Example – 11:

Find the value of 4 cos A + 3 cos B if angles A and B lies in second quadrants and

Solution:

sin A = 3/5 and sin B = 4/5

We have, sin²A + cos²A = 1

∴ (3/5)² + cos²A = 1

∴ cos²A = 1 – 9/25

∴  cos²A = (25 – 9)/25

∴  cos²A = 16/25

∴  cosA = ± 4/5

Now A lies in the second quadrant where cos ratio is negative.

∴  cosA = – 4/5

sin B = 4/5

We have, sin²B + cos²B = 1

∴ (4/5)² + cos²B = 1

∴ cos²B = 1 – 16/25

∴  cos²B = (25 – 16)/25

∴  cos²B = 9/25

∴  cosB = ± 3/5

Now A lies in the second quadrant where cos ratio is negative.

∴  cosA = – 3/5

The value of

4 cos A + 3 cos B = 4 x (- 4/5) + 3 x (- 3/5)= -16/5 – 9/5 = -25/5 = -1

Example – 12:

If 2 sin x = 1, π/2 < x < π and √2 cos y = 1, 3π/2 < y < 2π, find the value of

Solution:

2sin x = 1, hence sin x = 1/2, π/2 < x < π

We have, sin²x + cos²x = 1

∴ (1/2)² + cos²x = 1

∴ cos²x = 1 – 1/4

∴  cos²x = (4-1)/4

∴  cos²x = 3/4

∴  cos x = ± √3/2

Now x lies in the second quadrant where cos ratio is negative.

∴  cos x = – √3/2

Now, tan x = sin x/ cos x = (1/2)/(- √3/2) = – 1/√3

√2 cos y = 1, hence cos y = 1/√2,  3π/2 < y < 2π

We have, sin²y + cos²y = 1

∴ sin²y + (1/√2)² = 1

∴sin²y = 1 – 1/2

∴  sin²y = (2 – 1)/2

∴  sin²y = 1/2

∴  sin y = ±  1/√2

Now y lies in the third quadrant where sin ratio is negative.

∴  sin y = –  1/√2

Now, tan y = sin y/ cos y = ( –  1/√2)/(1/√2) = – 1

Example – 13:

If cos A = sin B = -1/3, where π/2 < A < π and  π < B< 3π/2, then find the value of

Solution:

Given cos A = -1/3, π/2 < x < π

We have, sin²A + cos²A = 1

∴ sin²A + (-1/3)² = 1

∴ sin²A = 1 – 1/9

∴  sin²A = (9-1)/9

∴  sin²A = 8/9

∴  sin A = ± 2√2/3

Now A lies in the second quadrant where sin ratio is positive.

∴  sin A = 2√2/3

Now, tan A = sin A/ cos A = (2√2/3) / (- 1/3)= – 2√2

sin B = – 1/3, π < B< 3π/2

We have, sin²B + cos²B = 1

∴(- 1/3)² + cos²B = 1

∴ cos²B = 1 – 1/9

∴  cos²B = (9 – 1)/9

∴  cos²B = 8/9

∴  cos B = ± 2√2/3

Now y lies in the second quadrant where cos ratio is negative.

∴  cos B = –  2√2/3

Now, tan B = sin B/ cos B = ( –  1/3)/(-  2√2/3) =  1/2√2

The post Signs of Trigonometric Ratios appeared first on The Fact Factor.

Let m∠ AOP = θ = 210^o = (7π/6)^c

m∠ POM = 30^o

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30^o)

OM = √3/2(OP) = √3/2 (1) = √3/2  (side opposite to 60^o)

Point P is in the third quadrant

Hence x = -√3/2 an y = -1/2. Thus

sin 210^o = y = – 1/2

cos 210^o =  x = – √3/2

tan 210^o =  y/x  = (-1/2)/(-√3/2) = 1/√3

cosec 210^o = 1/y = 1/(-1/2) = – 2

sec 210^o  =  1/x = 1/(-√3/2) = – 2/√3

cot 210^o = x/y = (-√3/2)/(-1/2) = √3

Trigonometric Ratios of 225^o or (5π/4)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 225^o = (5π/4)^c

m∠ POM = 45^o

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 45^o-45^o-90^o triangle

PM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

OM = 1/√2(OP) = 1/√2 (1) = √1/√2  (side opposite to 45^o)

Point P is in the third quadrant

Hence x = – 1/√2 an y = – 1/√2. Thus

sin 225^o = y = – 1/√2

cos 225^o =  x = – 1/√2

tan 22^o =  y/x  = (-1/√2)/(- 1/√2) =  1

cosec 225^o = 1/y = 1/(-1/√2) = – √2

sec 225^o  =  1/x = 1/(-1/√2) = – √2

cot 225^o = x/y = (-1/√2)/(-1/√2) = 1

Trigonometric Ratios of 240^o or (4π/3)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 240^o = (4π/3)^c

m∠ POM = 60^o

Let PM be perpendicular to OX’. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = √3/2(OP) = 1/2 (1) = √3/2  (side opposite to 60^o)

OM = 1/2(OP) = √3/2 (1) = 1/2  (side opposite to 30^o)

Point P is in the third quadrant

Hence x = – 1/2 an y = – √3/2. Thus

sin 240^o = y = – √3/2

cos 240^o =  x = -1/2

tan 240^o =  y/x  = (-√3/2)/(-1/2) = √3

cosec 240^o = 1/y = 1/(-√3/2) = – 2/√3

sec 240^o  =  1/x = 1/(-1/2) = – 2

cot 240^o = x/y = (-1/2)/(-√3/2) = 1/√3

Trigonometric Ratios of 270^o or (3π/2)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 270^o = (3π/2)^c

Let PM be perpendicular to OX. M coicides with O

PM = 1  and OM = 0

Point P is on negative y-axis

Hence x = 0 an y = -1. Thus

sin 270^o = y = -1

cos 270^o =  x = 0

tan 270^o =  y/x  (Not defined since x = 0)

cosec 270^o = 1/y = 1/-1 = -1

sec 270^o  =  1/x  (Not defined since x = 0)

cot 270^o = x/y = 0/-1 = 0

Trigonometric Ratios of 300^o or (5π/3)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 300^o = (5π/3)^c

Let PM be perpendicular to OX. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = √3/2(OP) = 1/2 (1) = √3/2  (side opposite to 60^o)

OM = 1/2(OP) = √3/2 (1) = 1/2  (side opposite to 30^o)

Point P is in the fourth quadrant

Hence x = 1/2 an y = – √3/2. Thus

sin 300^o = y = – √3/2

cos 300^o =  x = 1/2

tan 300^o =  y/x  = (-√3/2)/(1/2) = – √3

cosec 300^o = 1/y = 1/(- √3/2) = – 2/√3

sec 300^o  =  1/x = 1/(1/2) = 2

cot 300^o = x/y = (1/2)/(- √3/2) = – 1/√3

Trigonometric Ratios of 315^o or (7π/4)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 315^o = (7π/4)^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 45^o-45^o-90^o triangle

PM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

OM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

Point P is in the fourth quadrant

Hence x = 1/√2 an y = -1/√2. Thus

sin 315^o = y = – 1/√2

cos 315^o =  x = √1/√2

tan 315^o =  y/x  = (-1/√2)/(1/√2) = – 1

cosec 315^o = 1/y = 1/(-1/√2) = – √2

sec 315^o  =  1/x = 1/(1/√2) = √2

cot 315^o = x/y = (1/√2)/(-1/√2) = – 1

Trigonometric Ratios of 330^o or (11π/6)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 330^o = (11π/6)^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30^o)

OM = √3/2(OP) = √3/2 (1) = √3/2  (side opposite to 60^o)

Point P is in the fourth quadrant

Hence x = √3/2 an y = – 1/2. Thus

sin 330^o = y = – 1/2

cos 330^o =  x = √3/2

tan 330^o =  y/x  = (-1/2)/(√3/2) = – 1/√3

cosec 330^o = 1/y = 1/(-1/2) = – 2

sec 330^o  =  1/x = 1/(√3/2) = 2/√3

cot 330^o = x/y = (√3/2)/(-1/2) = – √3

Trigonometric Ratios of 360^o or 2π^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 360^o = 2π^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(1, 0)

Hence x = 1 an y = 0. Thus

sin 360^o = y = 0

cos 360^o =  x = 1

tan 360^o =  y/x  = 0/1 = 0

cosec 360^o = 1/y (Not defined since y = 0)

sec 360^o  =  1/x = 1/1 = 1

cot 360^o = x/y (Not defined since y = 0)

The post Trigonometric Ratios of Standard Angles in Third and Fourth Quadrants appeared first on The Fact Factor.

A circle with the centre at the origin and radius 1 is a standard unit circle. Let P(x, y) be any point on the unit circle with m∠ XOP = θ. Now P lies on the unit circle. Hence l(OP) = 1.

Let PM be perpendicular to OX. Thus ΔOMP is a right-angled triangle.

By Pythagoras theorem

OM² + MP² = OP²

x² + y² = 1

Then by definition of trigonometric ratios

sin θ = length of the opposite side / Length of the hypotenuse

sin θ = PM/ OP = y/1 = y

cos θ= length of the adjacent side / Length of the hypotenuse

cos θ= OM/ OP = x/1 = x

tan θ = length of opposite side / length of adjacent side

tan θ = PM/OM = y/x  (x not equal to 0)

cosec θ =  Length of hypotenuse / length of opposite side

cosec θ == OP/ PM = 1/y (y not equal to 0)

sec θ =  Length of hypotenuse / length of adjacent side

sec θ = OP/ OM = 1/x (x not equal to 0)

cot θ = length of adjacent side / length of opposite side

cot θ = OM/PM = x/y  (y not equal to 0)

From above values we can see that

cosec θ = 1/sin θ (if sin θ not equal to zero)

sec θ = 1/cos θ (if cos θ not equal to zero)

cot θ = 1/tan θ (if sin θ not equal to zero)

Notes:

• The trigonometric functions do not depend on the position of point P on the terminal ray but they depend on the measure of angle q.
• Coterminal angles have the same trigonometric functions
• Since x = cos θ and y = sinθ, point P has coordinates (cos θ, sin θ).

Let P(x, y) be on the standard unit circle such that

x² + y² = 1

x²  ≤  1

– 1 ≤ x ≤ 1

– 1 ≤ cos θ ≤ 1

Similarly

y²  ≤  1

– 1 ≤ y ≤ 1

– 1 ≤ sin θ ≤ 1

Similarly

sec θ ≥ 1 or sec θ ≤ – 1

cosec θ ≥ 1 or cosec θ ≤ – 1

tan θ and cot θ can be any real numbers.

Trigonometric Ratios of 0^o or 0^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 0^o = 0^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(1, 0)

Hence x = 1 an y = 0. Thus

sin 0^o = y = 0

cos 0^o =  x = 1

tan 0^o =  y/x  = 0/1 = 0

cosec 0^o = 1/y (Not defined since y = 0)

sec 0^o  =  1/x = 1/1 = 1

cot 0^o = x/y (Not defined since y = 0)

Trigonometric Ratios of 30^o or (π/6)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 30^o = (π/6)^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30^o)

OM = √3/2(OP) = √3/2 (1) = √3/2  (side opposite to 60^o)

Point P is in the first quadrant

Hence x = √3/2 an y = 1/2. Thus

sin 30^o = y = 1/2

cos 30^o =  x = √3/2

tan 30^o =  y/x  = (1/2)/(√3/2) = 1/√3

cosec 30^o = 1/y = 1/(1/2) = 2

sec 30^o  =  1/x = 1/(√3/2) = 2/√3

cot 30^o = x/y = (√3/2)/(1/2) = √3

Trigonometric Ratios of 45^o or (π/4)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 45^o = (π/4)^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX. Thus ΔOMP is 45^o-45^o-90^o triangle

PM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

OM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

Point P is in the first quadrant

Hence x = 1/√2 an y = 1/√2. Thus

sin 45^o = y = 1/√2

cos 45^o =  x = √1/√2

tan 45^o =  y/x  = (1/√2)/(1/√2) = 1

cosec 45^o = 1/y = 1/(1/√2) = √2

sec 45^o  =  1/x = 1/(1/√2) = √2

cot 45^o = x/y = (1/√2)/(1/√2) = 1

Trigonometric Ratios of 60^o or (π/3)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 60^o = (π/3)^c

Let PM be perpendicular to OX. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = √3/2(OP) = 1/2 (1) = √3/2  (side opposite to 60^o)

OM = 1/2(OP) = √3/2 (1) = 1/2  (side opposite to 30^o)

Point P is in the first quadrant

Hence x = 1/2 an y = √3/2. Thus

sin 60^o = y = √3/2

cos 60^o =  x = 1/2

tan 60^o =  y/x  = (√3/2)/(1/2) = √3

cosec 60^o = 1/y = 1/(√3/2) = 2/√3

sec 60^o  =  1/x = 1/(1/2) = 2

cot 60^o = x/y = (1/2)/(√3/2) = 1/√3

Trigonometric Ratios of 90^o or (π/2)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 90^o = (π/2)^c

Let PM be perpendicular to OX. M coicides with O

PM = 1  and OM = 0

Point P is on postive y-axis

Hence x = 0 an y = 1. Thus

sin 90^o = y = 1

cos 90^o =  x = 0

tan 90^o =  y/x  (Not defined since x = 0)

cosec 90^o = 1/y = 1/1 = 1

sec 90^o  =  1/x  (Not defined since x = 0)

cot 90^o = x/y = 0/1 = 0

Trigonometric Ratios of 120^o or (2π/3)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 120^o = (2π/3)^c

m∠ POM = 60^o

Let PM be perpendicular to OX’. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = √3/2(OP) = 1/2 (1) = √3/2  (side opposite to 60^o)

OM = 1/2(OP) = √3/2 (1) = 1/2  (side opposite to 30^o)

Point P is in the second quadrant

Hence x = – 1/2 an y = √3/2. Thus

sin 120^o = y = √3/2

cos 120^o =  x = -1/2

tan 120^o =  y/x  = (√3/2)/(-1/2) = -√3

cosec 120^o = 1/y = 1/(√3/2) = 2/√3

sec 120^o  =  1/x = 1/(-1/2) = – 2

cot 120^o = x/y = (-1/2)/(√3/2) = – 1/√3

Trigonometric Ratios of 135^o or (3π/4)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 135^o = (3π/4)^c

m∠ POM = 45^o

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 45^o-45^o-90^o triangle

PM = 1/√2(OP) = 1/√2 (1) = 1/√2  (side opposite to 45^o)

OM = 1/√2(OP) = 1/√2 (1) = √1/√2  (side opposite to 45^o)

Point P is in the second quadrant

Hence x = – 1/√2 an y = 1/√2. Thus

sin 135^o = y = 1/√2

cos 135^o =  x = – 1/√2

tan 135^o =  y/x  = (-1/√2)/(1/√2) = – 1

cosec 135^o = 1/y = 1/(1/√2) = √2

sec 135^o  =  1/x = 1/(-1/√2) = – √2

cot 135^o = x/y = (-1/√2)/(1/√2) = – 1

Trigonometric Ratios of 150^o or (5π/3)^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 150^o = (5π/6)^c

m∠ POM = 30^o

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(x, y)

Let PM be perpendicular to OX’. Thus ΔOMP is 30^o-60^o-90^o triangle

PM = 1/2(OP) = 1/2 (1) = 1/2  (side opposite to 30^o)

OM = √3/2(OP) = √3/2 (1) = √3/2  (side opposite to 60^o)

Point P is in the second quadrant

Hence x = – √3/2 an y = 1/2. Thus

sin 150^o = y = 1/2

cos 150^o =  x = – √3/2

tan 150^o =  y/x  = (1/2)/(-√3/2) = – 1/√3

cosec 150^o = 1/y = 1/(1/2) = 2

sec 150^o  =  1/x = 1/(-√3/2) = – 2/√3

cot 150^o = x/y = (-√3/2)/(1/2) = – √3

Trigonometric Ratios of 180^o or π^c:

Let us consider a standard unit circle

Let m∠ AOP = θ = 180^o = π^c

Ray OA is the initial arm of the angle.

The terminal arm of the angle ray OP intersects the circle at P(-1, 0)

Hence x = -1 an y = 0. Thus

sin 180^o = y = 0

cos 180^o =  x = – 1

tan 180^o =  y/x  = 0/-1 = 0

cosec180^o = 1/y (Not defined since y = 0)

sec 180^o  = 1/x = 1/-1 = – 1

cot 180^o = x/y (Not defined since y = 0)

The post Trigonometric Ratios of Standard Angles in First and Second Quadrants appeared first on The Fact Factor.

In this article, we shall study to solve problems based on the area of the sector.

Example – 01:

Find the area of a sector of the circle which subtends an angle of 120° at the centre, if the radius of the circle is 6 cm.

Given: Angle subtended at centre = θ = 120° = 120 x (π/180) = (2π/3)^c , Radius of circle = r = 6 cm.

To find: Area of sector = A =?

Solution:

Area of sector = ½ r²θ= ½ x 6² x (2π/3) =12π sq. cm

Ans: The area of the sector is 12π sq. cm

Example – 02:

The area of the circle is 81π sq. cm. Find the length of its arc subtending an angle of 150° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 81π sq. cm, Angle subtended at centre = θ = 150° = 150 x (π/180) = (5π/6)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Solution:

Area of circle = πr²  = 81π

∴  r²  = 81

∴  r  = 9 cm

Length of arc = S = r θ = 9 x (5π/6) = 7.5 π cm

Area of sector = ½ r²θ= ½ x 9² x (5π/6) = 33.75 π sq. cm

Ans: length of the arc is 7.5 π cm and the area of the sector is 33.75 π sq. cm

Example – 03:

The area of a circle is 25π sq. cm. Find the length of its arc subtending an angle of 144° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 25π sq. cm, Angle subtended at centre = θ = 144° = 144 x (π/180) = (4π/5)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr²  = 25π

∴  r²  = 25

∴  r  = 5 cm

Length of arc = S = r θ = 5 x (4π/5) = 4π cm

Area of sector = ½ r²θ= ½ x 5² x (4π/5) = 10π sq. cm

Ans: length of the arc is 4π cm and the area of the sector is 10π sq. cm

Example – 04:

The area of a circle is 81π sq. cm. Find the length of its arc subtending an angle of 300° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 81π sq. cm, Angle subtended at centre = θ = 300° = 300 x (π/180) = (5π/3)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr²  = 81π

∴  r²  = 81

∴  r  = 9 cm

Length of arc = S = r θ = 9 x (5π/3) = 15π cm

Area of sector = ½ r²θ= ½ x 9² x (5π/3) = 67.5 π sq. cm

Ans: length of the arc is 15π cm and the area of the sector is 67.5π sq. cm

Example – 05:

The perimeter of a sector of a circle of area 25π sq. cm is 20 cm. Find the area of the sector.

Given: Area of circle = 25π sq. cm, Perimeter = 20 cm

To find: Area of sector = A = ?

Area of circle = πr²  = 25π

∴  r²  = 25

∴  r  = 5 cm

Perimeter of sector = r + r + s = 20

∴ 2r + r θ = 20

∴ r (2 + θ) = 20

∴ 5 (2 + θ) = 20

∴  2 + θ = 4

∴   θ = 2^c

Area of sector = ½ r²θ= ½ x 5² x 2 = 25 sq. cm

Ans: The area of the sector is 25 sq. cm

Example – 06:

The perimeter of a sector of a circle of area 64π sq. cm is 56 cm. Find the area of the sector.

Given: Area of circle = 64π sq. cm, Perimeter = 56 cm

To find: Area of sector = A = ?

Area of circle = πr²  = 64π

∴  r²  = 64

∴  r  = 8 cm

Perimeter of sector = r + r + s = 56

∴ 2r + r θ = 56

∴ r (2 + θ) = 56

∴ 8 (2 + θ) = 56

∴  2 + θ = 7

∴   θ = 5^c

Area of sector = ½ r²θ= ½ x 8² x 5 = 160 sq. cm

Ans: The area of the sector is 160 sq. cm

Example – 07:

Find the area of sector whose arc length is 30π cm and the angle of the sector is 40°.

Given: Length of arc = 30π cm, angle of sector = θ = 40° = 40 x π/180 = (2π/9)^c ,

To find: Area of sector = A = ?

Length of arc = S =  r θ

∴ 30π =  r x (2π/9)

∴ r   = 135 cm

Area of sector = ½ r²θ= ½ x 135² x (2π/9) = 2025π sq. cm

Ans: The area of the sector is 2025 sq. cm

Example – 08:

In a circle of radius 12 cm, an arc PQ subtends the angle of 30° at the centre. Find the area between arc PQ and chord PQ.

Given: radius of circle = r = 12 cm, angle subtended at the centre = θ = 30° = 30 x (π/180) = (π/6)^c ,

To find: the area between arc PQ and chord PQ.

Solution:

Area of sector  = ½ r²θ= ½ x 12² x (π/6) = 12π sq. cm

In Δ OOR, sin 30° = QR/OQ

∴  OR = OQ sin 30° = 12 x 1/2 = 6 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6 = 36 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 12π – 36 = 12(π – 3) sq. cm

Ans: The area between arc PQ and chord PQ is 12(π – 3) sq. cm

Example – 09:

OPQ is the sector of a circle with centre O and radius 12 cm. if m ∠ POQ= 60°, find the difference between the areas of sector POQ and Δ POQ.

Given: radius of circle = r = 12 cm, angle subtended at the centre = θ = 60° = 60 x (π/180) = (π/3)^c ,

To find: the difference between the areas of sector POQ and Δ POQ.

Solution:

Area of sector  = ½ r²θ= ½ x 12² x (π/3) = 24π sq. cm

In Δ OQR, sin 60° = QR/OQ

∴  OR = OQ sin 60° = 12 x √3 /2 = 6√3 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6√3 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 24π – 36√3 = 12(2π – 3√3) sq. cm

Ans: The difference between the areas of sector POQ and Δ POQ. is 12(2π – 3√3) sq. cm

Example – 10:

OPQ is a sector of a circle with centre O and radius 12 cm. if m∠OPQ = 30°, Find the area between arc PQ and chord PQ.

Given: radius of circle = r = 12 cm,

To find: the area between arc PQ and chord PQ.

Solution:

Δ OPQ is isosceles triangle

m∠ OPQ = m∠ OQP =  30°

m∠ POQ = θ = 120° = 120 x π/180 = (2π/3)^c

Area of sector  = ½ r²θ= ½ x 12² x (2π/3) = 48π sq. cm

Δ OQR is 30°-60°-90° triangle

OR = ½OQ = ½ x 12 = 6 cm

QR = √3 /2 OQ = √3 /2 x 12 = 6√3 

PQ = 2 QR = 2 x 6√3  = 12√3 

Area of Δ POQ = ½ x base x height = ½ x PQ x OR = ½ x 12√3  x 6 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 48π – 36√3 = 12(4π – 3√3) sq. cm

Ans: The area between arc PQ and chord PQ. is 12(4π – 3√3) sq. cm

Example – 11:

Two circles each of radius 7 cm intersect each other such that the distance between their centres is 7√2 cm. Find area common to both the circles.

Given: radius of circle = r = 7 cm, Distance between centres = 7√2 cm

To find: the area of common portion = ?

Solution:

In quadrilateral ADBC

AD = DB = BC = CA = 7cm

diagonal AB = 7√2 cm

Hence quadrilateral ADBC is a square with each angle 90°

This is central angle subtended for sectors of both the circles = θ = 90° = 90 x π/180 = (π/2)^c

Area of common region = area of sector (A-CED) + Area of sector (B-CFD) – area of square ADBC

∴ Area of common region = ½ r²θ  + ½ r²θ  – r²

∴ Area of common region = r² θ  – r²

∴ Area of common region = r² ( θ – 1)

∴ Area of common region = 7² ( π/2 – 1)

∴ Area of common region = 49(π/2 – 1) sq. cm

Ans: The area common to both the circle is 49(π/2 – 1) sq. cm

Science > Mathematics > Trigonometry > Angle Measurement > Area of Sector

The post Area of Sector appeared first on The Fact Factor.

In this article, we shall solve problems based on the length of an arc (arc length).

Example – 01:

Find the length of the arc of a circle of diameter 10 cm, if the arc is subtending an angle of 36° at the centre.

Given: Diameter = 10 cm, radius = r = 10/2 = 5 cm, angle subtended = 36° = 36 x (π/180) = (π//5)^c

To Find: Length of arc = S = ?

Solution:

length of arc is given by

S = r θ  = 5 x (π//5) = π cm

Ans: The length of the arc is π cm

Example – 02:

Find the length of the arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.

Given: radius = r = 15 cm, angle subtended = 108° = 108 x (π/180) = (3π//5)^c

To Find: Length of arc = S = ?

Solution:

length of arc is given by

S = r θ  = 15 x (3π//5) = 9π cm

Ans: The length of the arc is 9π cm

Example – 03:

The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to the length of the radius.

Given: radius = r = 9 cm, length of chord = r

To Find: Length of arc = S = ?

Solution:

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x (π/180) = (π//3)^c

We have

S = r θ  = 9 x π//3 = 3π cm

Ans: The length of the arc is 3π cm

Example – 04:

In a circle of diameter 40 cm, the length of the chord is 20 cm. Find the length of the minor arch of the chord.

Given: diameter = 4o cm, radius = r = 20 cm, length of chord = 20 cm

To Find: Length of arc = S = ?

Solution:

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x π/180 = (π//3)^c

We have

S = r θ  = 20 x π//3 = 20π/3 cm

Ans: The length of the minor arc is 20π/3 cm

Example – 05:

A pendulum of 14 cm long oscillates through an angle of 18°. Find the length of the path described by its extremity.

Given: radius = r = 14 cm, angle subtended = 18° = 18 x (π/180) = (π//10)^c

To Find: Length of arc = S = ?

Solution:

We have

S = r θ  = 14 x π/10 = 1.4π cm

Ans: The length of arc is 1.4π cm

Example – 06:

Find in radians and degrees the angle subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of circle is 25 cm.

Given: radius = r = 25 cm, Length of arc = 15 cm

To Find: angle subtended at the centre = θ = ?

Solution:

We have

S = r θ

∴  θ = S/r = 15/25 = (3/5)^c

(3/5)^c = (3/5) x (180/π) = (108/π)°

Ans: The angle subtended at the centre is (3/5)^c or (108/π)°

Example – 07:

Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. Take ( π = 22/7)

Given: Length of arc = S = 37.4 cm, angle subtended = 60° = 60 x π/180 = (π//3)^c

To Find: Length of arc = S = ?

Solution:

We have

S = r θ

∴   r =S/θ = 37.4/(π//3) = (37.4 x 3) / (22/7)

∴   r = (37.4 x 3 x 7) / 22 = 35.7

Ans: The length of the arc is 35.7 cm

Example – 08:

A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?

Given: radius = r = 4 cm, length of wire = length of arc = 10 cm

To Find: angle subtended at the centre = θ = ?

Solution:

We have

S = r θ

∴  θ = S/r = 10/4 = (2.5)^c

(2.5)^c = (2.5) x (180/π) = (450/π)

Ans: The angle subtended at the centre is (2.5)^c or (450/π)°

Example – 09:

Δ PQR is an equilateral triangle with a side 18 cm. A circle is drawn on segment QR as a diameter. Find the length of the arc of this circle intercepted within the triangle.

Given: Side of equilateral triangle = 18 cm, radius = 18/2 = 9 cm

To Find: length of the arc of circle intercepted = S =?

Solution:

s Δ PQR is an equilateral triangle, its each angle is 60°

Hence the triangles Δ QOE and Δ ROD are also equilateral triangles

∠ EOQ = ∠ ROD = 60°

EOD = 60° =  θ

Now, we have

S = r θ  = 9 x π//3 = 3π cm

Ans: The length of arc is 3π cm

Example – 10:

Two arcs of the same length subtend angle 60° and 75° at the centres of the circles. What is the ratio of the radii of the two circles?

Given: Angles subtended, θ₁ = 60° and θ₂ = 75°

To Find: Ratio of radii = r₁/r₂ = ?

Solution:

length of arc is given by S = r θ

For the first arc S₁ = r₁θ₁  ………. (1)

For the second arc S₂ = r₂θ₂  ………. (2)

Now length of two arcs is the same

S₁ = S₂

∴  r₁θ₁ =  r₂θ₂

∴  r₁/r₂ =  θ₂/ θ₁

∴  r₁/r₂ =  75°/ 60° = 5/4

Ans: The ratio of radii is 5:4

Example – 11:

Two arcs of the same length subtend angle 65° and 110° at the centres of the circles. What is the ratio of the radii of the two circles?

Given: Angles subtended, θ₁ = 65° and θ₂ = 110°

To Find: Ratio of radii = r₁/r₂ = ?

Solution:

length of arc is given by S = r θ

For the first arc S₁ = r₁θ₁  ………. (1)

For the second arc S₂ = r₂θ₂  ………. (2)

Now length of two arcs is the same

S₁ = S₂

∴  r₁θ₁ =  r₂θ₂

∴  r₁/r₂ =  θ₂/ θ₁

∴  r₁/r₂ =  110°/ 65° = 22/13

Ans: The ratio of radii is 22:13

Example – 12:

A train is running on a circular track of a radius of 1 km at a rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.

Given: Radius of the arc = r = 1 km = 1000 m, Speed of train = v = 36 km per hour = 36 x 1000/3600 = 10 m/s, time taken = t = 30 s.

To Find: Angle through which the train turns = θ = ?

Solution:

Distance covered by train  i.e. length of arc =  speed x time = 10 x 30 = 300 m

S = r θ

∴  θ = S/r = 300/1000 = (0.3)^c

∴  θ = (0.3)^c = (0.3) x (180/π) = (54/3.142)° = 17.19°

∴  θ = 17° + 0.19° = 17° + 0.19 x 60′ = 17° + 11′ = 17°,11′

Ans: The train will turn through 17°,11′

Example – 13:

A train is running on a circular track of a radius of 1500 m at the rate of 66 km per hour. Find the angle to the in radian, through which it will turn in 10 seconds.

Given: Radius of the arc = r = 1500 m, Speed of train = v = 66 km per hour = 66 x 1000/3600 = 55/3 m/s, time taken = t = 10 s.

To Find: Angle through which the train turns = θ = ?

Solution:

Distance covered by train  i.e. length of arc =  speed x time = (55/3) x 10 = 550/3 m

S = r θ

∴  θ = S/r = (550/3)/1500 = 550/4500 = (11/90)^c

Ans: The train will turn through (11/90)^c

Example – 14:

A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 m when it traces an angle of 72° at the centre, find the length of the rope. Take π = 22/7.

Given: central angle = θ = 72° = 72 x π/180 = (2π/5)^c, arc length = S = 88 m

To Find: Length of rope = r = ?

S = r θ

∴  r = S/θ = 88/(2π/5) = 220/π = 220 x 7/22 = 70 m

Ans: The length of the rope is 70 m.

Example – 15:

If the perimeter of a sector of a circle is four times the radius of the circle, find the central angle of the corresponding sector in radians.

Given:  Perimeter = 4 x radius = 4r

To find: Central angle = θ = ?

Perimeter of sector = r + r + s = 4r

∴ 2r + r θ = 4r

∴ r θ = 2r

∴   θ = 2^c

Ans: The central angle  is 2^c

Science > Mathematics > Trigonometry > Angle Measurement > Length of an Arc

The post Length of an Arc appeared first on The Fact Factor.

In this article, we shall study the problems based on the interior angles of a polygon, and the angle between the hour hand and a minute hand of a clock.

Example – 01:

If x^c = 405° and y° = – (π/12)^c. Find x and y

Solution:

Given x^c = 405°

x containing term on R.H.S. is in radians. Hence we should convert L.H.S. into radians

∴  x^c = 405° = 405 x π/180 = (9π/4)^c

∴  x = 9π/4

Given y° = – (π/12)^cy containing term on R.H.S. is in degrees. Hence we should convert L.H.S. into degrees

∴  y° = – (π/12)^c = – (π/12) x (180/π) = 15°

∴  y = 15

Ans: x = 9π/4 and y = 15

Example – 02:

If θ° = – (5π/9)^c and Φ^c = 900°. Find θ and Φ

Solution:

Given θ° = – (5π/9)^c

θ containing term on R.H.S. is in degrees. Hence we should convert L.H.S. into degrees

∴  θ° = – (5π/9)^c = – (5π/9) x (180/π) = – 100°

∴  θ = -100

Given Φ^c = 900°

Φ containing term on R.H.S. is in radians. Hence we should convert L.H.S. into radians

∴  Φ^c = 900° = 900 x π/180 = (5π)^c

∴  Φ = 5π

Ans: θ = -100 and Φ = 5π

Example – 03:

Express following angles in radians

– 35°45’30”

Solution:

– 35°45’30” = – [35° + (45/60)° + (30/3600)°]

– 35°45’30” = – [35° + 0.75° + 0.0083°] = – 35.7583°

– 35°45’30” = – 35.7583 x π/180 = 0.1987 π

– 35°45’30” = 0.1987 x 3.142 = 0.6242 radian

50°37’30”

Solution:

50°37’30” = 50° + (37/60)° + (30/3600)°

50°37’30” = 50° + 0.6167° + 0.0083° = 50.625°

50°37’30” = 50.625 x π/180 = 0.2812 π

50°37’30” = 0.2812 x 3.142 = 0.8837 radian

– 10°40’30”

Solution:

10°40’30” = 10° + (40/60)° + (30/3600)°

10°40’30” = 10° + 0.6667° + 0.0083° = 10.675°

10°40’30” = 10.675 x π/180 = 0.0593 π

10°40’30” = 0.0593 x 3.142 = 0.1863 radian

Interior Angle of Regular Polygon:

Steps to Find Interior Angle of Polygon:

1. Find the measure of each exterior angle of regular polygon = 360°/No.of sides of polygon
2. Find the measure of each interior angle of polygon = 180° – measure of exterior angle

Example – 04:

Find Interior angles of following regular polygons in degrees and radians

Pentagon:

Solution:

Pentagon has 5 sides

Each exterior angle  = 360°/5 = 72°

Each interior angle = 180° – 72° = 108° = 108 x π/180 = (3π/5)^c

Ans: The interior angle of a regular pentagon is 72° or (3π/5)^c

Hexagon:

Solution:

Hexagon has 6 sides

Each exterior angle  = 360°/6 = 60°

Each interior angle = 180° – 60° = 120° = 120 x π/180 = (2π/3)^c

Ans: The interior angle of a regular hexagon is 120° or (2π/3)^c

Octagon:

Solution:

Octagon has 8 sides

Each exterior angle  = 360°/8 = 45°

Each interior angle = 180° – 45° = 135° = 135 x π/180 = (3π/4)^c

Ans: The interior angle of a regular octagon is 135° or (3π/4)^c

A Polygon with 20 sides:

Solution:

Polygon has 20 sides

Each exterior angle  = 360°/20 = 18°

Each interior angle = 180° – 18° = 162° = 162 x π/180 = (9π/10)^c

Ans: The interior angle of a regular polygon with 20 sides is 162° or (9π/10)^c

A Polygon with 15 sides:

Solution:

Polygon has 15 sides

Each exterior angle  = 360°/15 = 24°

Each interior angle = 180° – 24° = 156° = 156 x π/180 = (13π/15)^c

Ans: The interior angle of a regular polygon with 15 sides is 156° or (13π/15)^c

A Polygon with 12 sides:

Polygon has 12 sides

Each exterior angle  = 360°/12 = 30°

Each interior angle = 180° – 30° = 150° = 150 x π/180 = (5π/6)^c

Ans: The interior angle of a regular polygon with 12 sides is 150° or (5π/6)^c

Example – 05:

Find the number of sides of polygon if each of its interior angle is (3π/4)^c.

Solution:

Each interior angle = (3π/4)^c = (3π/4) x (180/π) = 135°

Hence each exterior angle = 180° – 135° = 45°

Number of sides of polygon = 360°/each exterior angle = 360°/45 = 8°

Ans: Thus the polygon has 8 sides

Angle Between Hour Hand and Minute Hand:

Example – 06:

Find the degree and radian measure of the angle between the hour hand and minute hand of a clock at the following timings.

Twenty minutes past seven:

Solution:

At twenty minutes past seven, the minute hand is at 4 and hour hand crossed 7

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 20 minutes = 0.5° x 20 = 10°

Thus the hour hand is 10° ahead of 7 th Mark

The angle between 4 and 7 is 90°

Thus angle between hour hand and minute hand = 90° + 10° = 100°

100° = 100 x π/180 = (5π/9)^c

Ans: The angle between the hour hand and the minute hand is 100° or (5π/9)^c

Twenty minutes past two:

Solution:

At twenty minutes past two, the minute hand is at 4 and hour hand crossed 2

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 20 minutes = 0.5° x 20 = 10°

Thus the hour hand is 10° ahead of 2 nd Mark

The angle between 2 and 4 is 60°

Thus angle between hour hand and minute hand = 60° – 10° = 50°

50° = 50 x π/180 = (5π/18)^c

Ans: The angle between the hour hand and the minute hand is 50° or (5π/18)^c

Quarter past six:

Solution:

At quarter past six, the minute hand is at 3 and hour hand crossed 6

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 15 minutes = 0.5° x 15 = 7.5°

Thus the hour hand is 7.5° ahead of 6th Mark

The angle between 3 and 6 is 90°

Thus angle between hour hand and minute hand = 90° + 7.5° = 97.5°

97.5° = 97.5 x π/180 = (13π/24)^c

Ans: The angle between the hour hand and the minute hand is 97.5° or (13π/24)^c

Ten past eleven:

Solution:

At ten past eleven, the minute hand is at 2 and hour hand crossed 11

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 10 minutes = 0.5° x 10 = 5°

Thus the hour hand is 5° ahead of 11th Mark

The angle between 11 and 2 is 90°

Thus angle between hour hand and minute hand = 90° – 5° = 85°

85° = 85 x π/180 = (17π/36)^c

Ans: The angle between the hour hand and the minute hand is 85° or (17π/36)^c

Example – 07:

Show that the minute hand of a clock gains 5°30′ on hour hand in one minute.

Solution:

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by minute hand in 1 minute = 6°

Thus angle between hour hand and minute hand = 6° – 0.5° = 5.5° = 5°30′

Ans: Thus the minute hand of a clock gains 5°30′ on hour hand in one minute.

Example – 08:

Determine which of the following pairs of angles are coterminal.

210° and – 150°

– 150° = – 150° + 360° = 210°

Thus the two angles have the same initial arm and terminal arm.

Hence the angles 210° and – 150° are coterminal angles.

330° and – 60°

– 60° = – 60° + 360° = 300°

Thus the two angles do not have the same initial arm and terminal arm.

 Hence the angles 330° and – 60° are not coterminal angles.

405° and – 675°

405° = 405° – 360° = 45°

– 675° + 360° x 2 = 45°

Thus the two angles have the same initial arm and terminal arm.

 Hence the angles 405° and – 675° are coterminal angles.

1230° and – 930°

1230° = 1230° – 360° x 3 = 150°

– 930° + 360° x 3 = 150°

Thus the two angles have the same initial arm and terminal arm.

 Hence the angles 1230° and – 930° are coterminal angles.

Example – 09:

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in 1 second?

Given: No. of revolutions = 360 per minute

To Find: Radians per second =?

Solution:

No. of revolutions per second = 360/60 = 6

In one revolution the wheel turns through 2π radians

Radians per second = 2π x 6 = 12π^c

Ans: The wheel will turn through 12π^c in 1 second

Science > Mathematics > Trigonometry > Angle Measurement > Angle Measurement

The post Angle Measurement appeared first on The Fact Factor.

In this article, we shall study to find a measure of an angle in radians and degrees.

Express following angles in degrees, minutes, and seconds

Degrees x 60 = Minutes

Minutes x 60 = Seconds

• 74.87°

74.87° = 74° + 0.87°

∴  74.87° = 74° + 0.87 x 60′

∴  74.87° = 74° + 52.2′

∴  74.87° = 74° + 52′ + 0.2′

∴  74.87° = 74° + 52′ + 0.2 x 60”

∴  74.87° = 74° + 52′ + 12”

∴  74.87° = 74°,52′,12”

• – 30.6947°

– 30.6947° = – (30° + 0.6947°)

∴  – 30.6947° = – (30° + 0.6947 x 60′)

∴  – 30.6947° = – (30° + 41.682′)

∴  – 30.6947° = – (30° + 41′ + 0.682′)

∴  – 30.6947° = – (30° + 41′ + 0.682 x 60”)

∴  – 30.6947° = – (30° + 41′ + 40.92”)

∴  – 30.6947° = – (30° + 41′ + 41”)

∴  – 30.6947° = – 30°,41′,41” approx.

• 321.9°

321.9°  = 321° + 0.9°

∴  321.9° = 321° + 0.9 x 60′

∴  321.9° = 321° + 54′

∴  321.9° = 321°,54′,0”

• 200.6°

200.6°  = 200° + 0.6°

∴  200.6° = 200° + 0.6 x 60′

∴  200.6° = 200° + 36′

∴  200.6° = 200°,36′,0”

• 11.0133°

11.0133° = 11° + 0.0133°

∴  11.0133° =  11° + 0.0133 x 60′

∴ 11.0133° =  11° + 0.798′

∴  11.0133° =  11° + 0′ + 0.798′

∴  11.0133° =  11° + 0′ + 0.798 x 60”

∴  11.0133° =  11° + 0′ +  47.88”

∴  11.0133° =  11° + 0′ + 48”

∴ 11.0133° = 11°,48” approx.

• 94.3366°

94.3366° = 94° + 0.3366°

∴  94.3366° =  94° + 0.3366 x 60′

∴ 94.3366° =  94° + 20.196′

∴  94.3366° =  94° + 20′ + 0.196′

∴  94.3366° =  94° + 20′ + 0.196 x 60”

∴  94.3366° =  94° + 20′ +  11.76”

∴  94.3366° =  94° + 20′ + 12”

∴ 11.0133° = 94°,20′,12”  approx.

Conversion of Angles in Degrees into Radians

Degrees x π/180 = Radians

Conversion of Angles in Radians into Degrees

Radians  x 180/π = Degrees

Problems Based on Degree and Radian Measures of Angles:

Example – 01:

The difference between the two acute angles of a right-angle triangle is (2π/5)^c.  Find the angles in degrees.

Solution:

Let the two acute angles be x and y in degrees

Given their difference is  (2π/5)^c.=  (2π/5) x  (180/π) = 72°

∴  x – y = 72° ………   (1)

Now the sum of acute angles of triangle is always 90°

∴  x + y = 90° ………   (2)

Solving equation (1) and (2) we get

x = 81° and y = 9°

Ans: The acute angles of triangle are 81° and 9°

Example – 02:

The difference between the two acute angles of a right-angled triangle is (3π/10)^c.  Find the angles in degrees.

Solution:

Let the two acute angles be x and y in degrees

Given their difference is  (3π/10)^c.=  (3π/10) x  (180/π) = 54°

∴  x – y = 54° ………   (1)

Now the sum of acute angles of triangle is always 90°

∴  x + y = 90° ………   (2)

Solving equation (1) and (2) we get

x = 72° and y = 18°

Ans: The acute angles of triangle are 72° and 18°

Example – 03:

The sum of the two angles is 5π^c and their difference is 60°. Find the angles in degrees.

Solution:

Let the two acute angles be x and y in degrees

Given their sum is 5π^c.=  (5π) x  (180/π) = 900°

∴  x + y = 900° ………   (1)

Given their differene is 60°

∴  x – y = 60° ………   (2)

Solving equation (1) and (2) we get

x = 480° and y = 420°

Ans: The acute angles of triangle are 480° and 420°

Example – 04:

The measures of angles of a triangle are in the ratio 2:3:5. Find their measures in radians.

Solution:

The  angles.of triangle are in the ratio 2:3:5

Let the three angles be 2k, 3k, and 5k.

Now the sum of all angles of triangle is 180°

∴  2k + 3k + 5k = 180°

∴  10k   = 180°

∴  k   = 18°

∴ The three angles are ( 2 x 18° = 36°), (3 x 18° = 54°), and (5 x 18°) = 90°

36° = 36 x π/180 = (π/5)^c

54° = 54 x π/180 = (3π/10)^c

90° = 90 x π/180 = (π/2)^c

Ans: The angles of quadrilateral are (π/5)^c, (3π/10)^c, and (π/2)^c,

Example – 05:

One angle of a triangle is (2π/9)c and the measures of the other two angles.are in the ratio 4:3. Find their measures in degrees and radians.

Solution:

one of the angle of triangle is of measure (2π/9)^c .=  (2π/9) x  (180/π) = 40°

other two angles.are in the ratio 4:3

Let the two angles be 4k, and 3k.

Now the sum of all angles of triangle is 180°

∴  4k + 3k + 40°  = 180°

∴  7k   = 140°

∴  k   = 20°

∴ The two angles are ( 4 x 20° = 80°) and  (3 x 20° = 60°)

80° = 80 x π/180 = (4π/9)^c

60° = 60 x π/180 = (π/3)^c

Ans: The  two angles of triangle are 80° and 60° or (4π/9)^c and (π/3)^c,

Example – 06:

In ΔABC, m∠A = (2π/3)^c and m∠B = 45°. Find m∠C in both the system.

Solution:

m∠A = (2π/3)^c =  (2π/3) x  (180/π) = 120°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  120° +  45° + m∠C = 180°

∴  m∠C = 180° – 165°

∴  m∠C = 15° = 15 x (π /180) = (π/12)^c

Ans:  m∠C = 15° or (π/12)^c

Example – 07:

If the radian measures of two angles of a triangle are (5π/9)^c and (5π/18)^c. Find the measure of the third angle in radians and degrees.

Solution:

Let m∠A = (5π/9)^c =  (5π/9) x  (180/π) = 100°

Let m∠B = (5π/18)^c =  (5π/18) x  (180/π) = 50°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  100° +  50° + m∠C = 180°

∴  m∠C = 180° – 150°

∴  m∠C = 30° = 30 x (π /180) = (π/6)^c

Ans: Measure of third angle is (π/6)^c or  30°

Example – 08:

If the radian measures of two angles of a triangle are (3π/5)^c and (4π/15)^c. Find the measure of the third angle in radians and degrees.

Solution:

Let m∠A = (3π/5)^c =  (3π/5) x  (180/π) = 108°

Let m∠B = (4π/15)^c =  (4π/15) x  (180/π) = 48°

Now the sum of all angles of triangle is 180°

m∠A + m∠B + m∠C = 180°

∴  108° +  48° + m∠C = 180°

∴  m∠C = 180° – 156°

∴  m∠C = 24° = 24 x (π /180) = (2π/15)^c

Ans: Measure of third angle is (2π/15)^c or  24°

Example – 09:

In ΔLMN, m∠L = (3π/4)^c and m∠N = 30°. Find m∠M in both the system.

Solution:

m∠L = (3π/4)^c =  (3π/4) x  (180/π) = 135°

Now the sum of all angles of triangle is 180°

m∠L + m∠M + m∠N = 180°

∴  135° +  m∠M + 30° = 180°

∴  m∠M = 180° – 165°

∴  m∠M = 15° = 15 x (π /180) = (π/12)^c

Ans:  m∠M = 15° or  (π/12)^c

Example – 10:

One angle of a quadrilateral is (2π/9)c and the measures of the other three angles.are in the ratio 3:5:8. Find their measures in radians.

Solution:

one of the angle of quadrilateral is of measure (2π/9)^c .

=  (2π/9) x  (180/π) = 40°

other three angles.are in the ratio 3:5:8

Let the three angles be 3k, 5k, and 8k.

Now the sum of all angles of quadrilateral is 360°

∴  3k + 5k + 8k + 40°  = 360°

∴  16k   = 320°

∴  k   = 20°

∴ The three angles are ( 3 x 20° = 60°), (5 x 20° = 100°), and (8 x 20°) = 160°

60° = 60 x π/180 = (π/3)^c

100° = 100 x π/180 = (5π/9)^c

160° = 160 x π/180 = (8π/9)^c

Ans: The angles of quadrilateral are (2π/9)^c, (π/3)^c, (5π/9)^c, and (8π/9)^c,

Example – 11:

One angle of a quadrilateral is (2π/5)c and the measures of the other three angles.are in the ratio 2:3:4. Find their measures in degrees and radians.

Solution:

one of the angle of quadrilateral is of measure (2π/5)^c .

=  (2π/5) x  (180/π) = 72°

other three angles.are in the ratio 2:3:4

Let the three angles be 2k, 3k, and 4k.

Now the sum of all angles of quadrilateral is 360°

∴  2k + 3k + 4k + 72°  = 360°

∴  9k   = 288°

∴  k   = 32°

∴ The three angles are ( 2 x 32° = 64°), (3 x 32° = 96°), and (4 x 32°) = 128°

64° = 64 x π/180 = (16π/45)^c

96° = 96 x π/180 = (24π/45)^c

128° = 108 x π/180 = (32π/4)^c

Ans: The angles of quadrilateral are (2π/9)^c, (π/3)^c, (5π/9)^c, and (8π/9)^c,

Example – 12:

The measures of angles of a quadrilateral are in the ratio 2:3:6:7. Find their measures in degrees and radians.

Solution:

The measures of angles are in the ratio 2:3:6:7

Let the measures of angles be 2k, 3k, 6k  and 7k.

Now the sum of all angles of quadrilateral is 360°

∴  2k + 3k + 6k + 7k  = 360°

∴  18k   = 360°

∴  k   = 20

∴ The measures of angles are ( 2 x 20° = 40°), (3 x 20° = 60°), (6 x 20° = 120°), and (7 x 20°) = 140°

40° = 40 x π/180 = (2π/9)^c

60° = 60 x π/180 = (π/3)^c

120° = 120 x π/180 = (2π/3)^c

140° = 140 x π/180 = (7π/9)^c

Ans: The measures of angles of quadrilateral are 40°, 60°, 120°, and 140°

or (2π/9)^c, (π/3)^c, (2π/3)^c, and (7π/9)^c,

Example – 13:

The measures of angles of a quadrilateral are in the ratio 3:4:5:6. Find their measures in degrees and radians.

Solution:

The measures of angles are in the ratio 3:4:5:

Let the measures of angles be 3k, 4k, 5k  and 6k.

Now the sum of all angles of quadrilateral is 360°

∴  3k + 4k + 5k + 6k  = 360°

∴  18k   = 360°

∴  k   = 20

∴ The measures of angles are ( 3 x 20° = 60°), (4 x 20° = 80°), (5 x 20° = 100°), and (6 x 20°) = 120°

60° = 60 x π/180 = (π/3)^c

80° = 80 x π/180 = (2π/9)^c

100° = 100 x π/180 = (5π/9)^c

120° = 120 x π/180 = (2π/3)^c

Ans: The measures of angles of quadrilateral are 60°, 80°, 100°, and 120°

or (π/3)^c, (2π/9)^c, (5π/9)^c, and (2π/3)^c,

Example – 14:

The angles of triangle are in A.P. and the greatest angle is 84°. Find all the three angles in radians.

Solution:

Let the three angles of a triangle in A.P. be (a – d), a, (a + d) in degrees

Now the sum of all angles of a triangle is 180°

(a – d) + a + (a + d) = 180°

∴ 3a = 180°

∴ a = 60° = 60 x π/180 = (π/3)^c

Now the greatest angle is 84°

∴ a + d = 84°

∴ 60° + d = 84° = 84 x π/180 = (7π/15)^c

∴ d = 24°

∴ a – d = 60° – 24° = 36° = 36 x π/180 = (π/5)^c

Ans: Measure of the angleare of triangle are (π/5)^c, (π/3)^c, (7π/15)^c,

Example – 15:

The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Express the least angle in radians.

Solution:

Let the four angles of quadrilateral in A.P. be (a – 3d), (a -d), (a +d), and (a + 3d) in degrees

Now the sum of all angles of quadrilateral is 360°

(a- 3d) + (a – d) + (a + d) + (a + 3d) = 360°

∴ 4a = 360°

∴ a = 90°

Now the greatest angle is double the least

∴ a + 3d = 2(a – 3d)

∴ 90 + 3d = 2(90 – 3d)

∴ 90 + 3d = 180 – 6d

∴ 9d = 90

∴ d = 10°

Least angle = a – 3d = 90° – 3 x 10° = 90° – 30° = 60° = 60 x (π /180) = (π/3)^c

Ans: The least angle in radians is (π/3)^c.

Science > Mathematics > Trigonometry > Angle Measurement > Measurement of Angle in Radians

The post Measurement of Angle in Radians appeared first on The Fact Factor.