In the last article, we have studied measurement of distances between heavenly bodies. In this article, we shall study methods of the mesurements of terrestrial distances.

An instrument used for Measurement of Bearings:

An instrument used for Measurement of Angle of Elevation and Depression:

Terminology Used in Measurement of Distances:

Angle of Elevation:

Suppose we are looking through a binocular or a telescope which is pointed horizontally and if we tilt the binocular or telescope in a vertical plane so that it points upward to observe a top of a tower or flying kite. etc. Then the angle through which we have tilted the binocular or telescope upward is called the angle of elevation of that object.

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer’s eye (the line of sight or LOS).

Angle of Depression:

Suppose we are looking through a binocular or a telescope which is pointed horizontally and if we tilt the binocular or telescope in a vertical plane so that it points downward to observe a bottom of a tower or a ship from a lighthouse. etc. Then the angle through which we have tilted the binocular or telescope downward is called the angle of elevation of that object.

Compass Bearing:

If O is the position of an observer and A is the object to be observed, then the acute angle that OA makes with the direction of a cardinal point on one side of it gives the compass bearing.

It means look north turn 63° towards the east to observe A along OA

Alternate representation is E27°N

It means look East turn 27° towards north to observe A along OA

True Bearing:

The angle made by the line joining the object to the observing point with the north measured clockwise from the north is called true bearing.

In the previous definition in the first case, the true bearing is 63° and in the second case, it is 216°.

Measurement of Height by Triangulation Method:

This method is used on a bright sunny day. On a sunny day, the tree (or the pole) cast a shadow.

Let AB be the tree (or the pole), where B be the base and BC be its shadow. Now take a reference pole ED of known length and keep it upright on the ground so that it is perpendicular to BC and shadow of its tip coincides with the shadow of the tip of the reference pole. The distances, BC and DC are measured.

Now ΔABC and ΔEDC are similar

Hence its corresponding sides are in proportions

AB / ED = BC / CD

By measuring ED, BC, and CD, the height of tree AB can be calculated.

Example – 01:

On a sunny day, a pole casts 30 m long shadow. An iron rod dug vertically also casts its shadow as far as that of the pole. If the height of the rod above the ground is 2 m and its dug is at 3 m from the tip of the shadow towards the pole, find the height of the pole.

AB = pole, DE = iron rod, BC = shadow of pole, DC = shadow of iron rod

BC = 30 m, DC = 3 m, ED = 2 m

Now ΔABC and ΔEDC are similar

Ans: The height of the pole is 20 m.

Measurement of Height by Trigonometric Method:

This method is used on a bright sunny day. On a sunny day, the tree (or the pole) cast a shadow.

Let AB be the tree (or the pole), where B be the base. Let us assume that the observer is at C. The distance BC is measured. The angle of elevation θ at point C for the top of the tree is measured. Using the trigonometric ratio length can be calculated as follows.

tan θ = AB/BC

∴  AB = BC tan θ

knowing BC and θ, the height of the tree can be calculated.

Example – 02:

The angle of elevation of the top of a tower at a distance of 150 m from its foot is 32°. Find the height of the tower.

AB = Tower, C = observing point

BC = 150 m

we have tanθ = AB/BC

∴  AB = 150 x  tan 32°

∴  AB = 150 x  0.6249 = 93.74 m

Ans: The height of the tower is 93.74 m or the height of the tower is 94 m significantly

Example – 03:

A student stands at a distance of 50 m from a foot of a school building and finds the angle of elevations for the top of the building and a top of flagpole mounted on the top of the building to be 60° and 63° respectively. Find the height of the building and the flagpole.

AB = Building, AD = Flagpole, C = point of observation

BC = 50 m

In Δ ABC, we have tan60° = AB/BC

∴  AB = 50 x  tan 60°

∴  AB = 50 x  1.732 = 86.6 m

In Δ DBC, we have tan63° = DB/BC

∴  DB = 50 x  tan 63°

∴  AB = 50 x  1.963 = 98.15 m

AD = DB – AB = 98.15 – 86.6 = 12.15 m

Ans: The height of school building is 86.6 m and height of flagpole is 12.15 m or The height of school building is 87 m and height of flagpole is 12 m significantly.

Example – 04:

A boy flying a kite has let out 60 m of string. If the angle of elevation of the kite is 60°, find the height of the kite above the ground. Assume there is no slack in the string due to its weight.

A = position of kite, AC = string, C = position of boy

AC = 60 m, θ = angle of elevation = 60°

sin θ = AB/AC

∴  AB = AC sin θ

∴  AB = 60 x sin 60° = 60 x 0.866 = 51.96 m

Ans: Height of the kite above ground is 51.96 m or the height of the kite above ground is 52 m significantly

Example – 05:

From the top of a lighthouse, 40.0 m above the water, the angle of depression of a small boat is 20°. Estimate the distance of the boat from the lighthouse.

AB = lighthouse, A = top of lighthouse and observation point, C = position of boat

AB = 40 m, θ = angle of depression = 20°

we have tanθ = AB/BC

∴  BC = AB/tan 20°

∴  BC = 40/ 0.3640 = 109.9 m

Ans: The distance of boat from the lighthouse is 109.9 m or the height of the tower is 110 m significantly

Example – 06:

the angle of elevation of the sun is 53°. Find the length of the shadow cast by a tree of height 20 m.

A = Top of tree, B = Foot of tree, BC = Shadow of tree

Position of sun is on line Ac in the sky

AB = 20 m, θ = angle of elevation = 53°

tan θ = AB/BC

∴  BC = AB/tan θ

∴  AB = 20 / tan 53° = 20 / 1.327 = 15.07 m

Ans: The length of the shadow is 15.07 m or the length of the shadow is 15 m significantly

Example – 07:

A motor car moves in a direction of 035° (N35°E) for 2 hours at 30 km/h. How far north and how far east does it travel? Express the answer correct up to 1 decimal.

Distance travelled = 30 x 2 = 60 km

True bearing = 035° (or compass bearing N35°E)

OA = Distance traveled along 035°

AB = Distance traveled towards the north

OB = Distance traveled towards the east

sinθ = AB/OA

∴  AB = OA sin θ

∴  AB = 60 x  sin 55° = 60 x 0.8192 = 49.15 = 49.2 km

cosθ = OB/OA

∴  OB = OA cos θ

∴  OB = 60 x  cos 55° = 60 x 0.5736 = 34.42 = 34.4 km

Ans: The motor car moved 49.2 km towards north and 34.4 km towards east.

Example – 08:

A soldier observes an enemy vehicle on a ground from the observation tower at height of 180 m from ground level. The angle of depression is 29°. Calculate the distance of the enemy vehicle from the foot of the tower. After some time. it is observed that after 1 minute the vehicle is 200 m from the foot of observation tower. Calculate the new angle of depression. Also, find the speed of the vehicle if it is moving in a straight line towards the observation tower.

Part – I:

AB =observation tower, A = observation point, C = position of vehicle

AB = 180 m, θ = angle of depression = 29°

we have tanθ = AB/BC

∴  BC = AB/tan 29°

∴  BC = 180/ 0.5543 = 324.7 m

Part – II:

AB =observation tower, A = observation point, C = position of vehicle

AB = 180 m, BC = 200 m

we have tanθ = AB/BC

∴  tanθ = 180/200 = 0.9

∴  θ = tan^-1(0.9) = 41°59′

Part-III

The distance travelled by vehicle in one minute = 324.7 – 200 = 124.7 m

Speed of vehicle = distance/ time = 124.7/60 = 2.01 m/s

Speed of vehicle = 2.01 x 18/5 = 7.24 km/h

Ans: The distance of vehicle from the observation tower was 324.7 m, the new angle of depression is 41°59′, and the speed of vehicle is 7.24 km/h

Measurement of Height of Inaccessible Mountain:

Let AB represents the height of the mountain and BD represents horizontal level. Let θ₁ and θ₂ be the angle of elevation of top (summit) of the mountain at point C and D measured by a sextant. distance CD can be measured directly.

In ΔABC cot θ₁ = BC/AB

In ΔABD cot θ₂ = BD/AB

∴   cot θ₁  –  cot θ₂  = BC/AB –  BD/AB

∴   cot θ₁  –  cot θ₂  = (BC –  BD) /AB

Measuring CD, θ₁, and θ₂ the height AB of the mountain can be found.

Example – 09:

The shadow of a pole standing on a level ground is found to be 45 m longer when the sun’s altitude is 30° than when it is 60° Determine height of the tower.

AB = Pole, θ₁= 60° and θ₂= 30°, CD = 45 m

Ans: Height of tower is 38.97 m

Example – 10:

The angle of elevation of the top of a tower at a certain distance was 220, on advancing a distance of 30 m towards the tower, the angle of elevation becomes 31o. Find the height of the tower.

AB = Pole, θ₁= 31° and θ₂= 22°, CD = 30 m

Ans: Height of tower is 36.99 m

Measurement of Width of River:

Let AB be the width of a river. An object like a tree at B is selected as a reference on the opposite bank of the river. A pole is fixed at point C on the bank where measurement is being done. Now proceeding along line BC, a pole is fixed at E such that Points B, C and E are on the same line.

Example – 11:

In the experiment to find the width of a river, measurements obtained are as shown in the figure. Calculate the width of the river.

We have AB = (DE/CD)x AB = (12/6)x 15 = 30 m

Ans: The width of river is 30 m.

Measurement of Aeroplane from Tracking Centre by RADAR Signal Echo Method:

RADAR is the abbreviation of RAdio Detection And Ranging. Electromagnetic energy pulses are beamed towards the object from a powerful transmitter. They travel at a speed of v = 3 x 10⁸ m/s and small wavelength. These waves are reflected (echo) back from the surface of the object and are received back at the position of RADAR.

The time interval (t) between the sending and receiving signals is measured accurately. The distance between the object and the position of RADAR is calculated using the formula

Distance = vt / 2

The same method can be used to determine the distance of the moon from the earth.

Measurement of Depth of Sea by SONAR Signal Echo Method:

SONAR is the abbreviation of SOund Navigation And Ranging. Ultrasound pulses are beamed towards the bed of the sea from a powerful transmitter. These waves (pulses) are reflected (echo) back from the surface of the bed pf sea and are received back at the receiver.

The time interval (t) between the sending and receiving signal is measured accurately. Knowing the speed (v) of the ultrasonic pulse in sea water, the depth of the sea at the place is calculated using the formula

Depth = vt / 2

Measurement of Distance of a Building  Wall From Observer by Echo Method:

Sound waves are beamed towards the building or the wall. These sound waves are reflected (echo) back from the surface of the building or wall and are listened as an echo by the observer.

The time interval (t) between the sending sound and receiving echo is measured accurately. Knowing the speed (v) of the sound at that place at that temperature, the distance is calculated using the formula

Distance = vt / 2

Measurement of thickness of a Metal Block by Echo Method:

Ultrasound pulses are beamed from one surface of the block using a powerful transmitter. These waves (pulses) travel towards the opposite surface and are reflected (echo) back fand are received back at the receiver.

The time interval (t) between the sending and receiving signals is measured accurately. Knowing the speed (v) of the ultrasonic pulse in the metal, the thickness of the metal block is calculated using the formula

ThicknessDepth = vt / 2

This method is also used to detect and position the defects in welding and castings.

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