## Height and Distance

Trigonometric ratios are used to solve the problems related to height and distance. This is one of the important application of trigonometry. It helps to find the heights of the objects like tree, building, tower etc. and distance between the objects.

While finding such height and distances, we have to solve the right angled triangle. For example, in the figure given below, AB is the height of a tree, BC is the distance between the tree and the observer at C.

Here, if θ and BC are known, we can calculate the height of tree AB, or if θ and AB are known, we can calculate the distance between the tree and the observer by using trigonometric ratio.

### Angle of elevation and depression

In the figure given below, an observer is looking at the top of a tower. Here, OA is called the line of sight and OB is the horizontal line. BOA is the angle between the upward line of sight and the horizontal line. Here, BOA is called the angle of elevation.

Thus, the angle made by the upward line of sight with horizontal line is known as angle of elevation.

And, in the figure given below, BOA is the angle between the downward line of sight and the horizontal line. BOA here, is called the angle of depression.

Thus, the angle made by downward line of sight with horizontal line is known as angle of depression.

In the figure given below, the horizontal lines AC and BO are parallel to each other. The angle of depression CAO and the angle of elevation BAO are alternate angles between parallel lines.

CAO = BAO, i.e. Angle of depression = Angle of elevation.

The instrument which is used to measure the angle of elevation or the angle of depression is called Theodolite.

### Worked Out Examples

Example 1: A man observes the top of a tower 40 m high and finds the elevation of 30°. Find the distance between the man and the foot of the tower.

Solution:

Let AB be the height of the tower and BC be the distance between the foot of the tower and the man.

Here, AB = 40 m

Angle of elevation, BCA = 30°

In right angled triangle ABC,

tan30° = p/b = AB/BC = 40/BC

or,     1/√3 = 40/BC

or,     BC = 40√3 = 40 × 1.732 = 69.28 m

So, the required distance is 69.28 m.

Example 2: An electric pole 20 m high is supported by a wire fixing its one end on the ground at some distance from the pole. If the wire joining the top of the pole is inclined to the ground at an angle of 30°, find the length of the wire.

Solution:

Let AB be the height of the pole and AC be the length of wire.

Here, AB = 20m.

Angle of elevation, BCA = 30°.

In right angled triangle ABC,

sin30° = p/b = AB/AC = 20/AC

or,     ½ = 20/AC

or,     AC = 40.

So, the required length of the wire is 40m.

Example 3: The circumference of a circular pond is 176m and a pillar is fixed at the centre of the pond. If a person finds the angle of elevation of 60° of the top of the pillar from any point on the bank of the pond, find the height of the pillar above the water pillar.

Solution:

Let AB be the height of the pillar above the  water level and BC be the radius of the pond.

Here, the circumference of the pond = 176m

i.e.     2πr = 176

or,     2 × 22/7 × BC = 176

or,     44/7 × BC = 176

or,     BC = 176 × 7/44

or,     BC = 28m

The angle of elevation, BCA = 60°

Now, in right angled triangle ABC,

tan60° = p/b = AB/BC = AB/28

or,     √3 = AB/28

or,     AB = 28√3 = 28 × 1.372 = 48.5

So, the required height of the pillar is 48.5m.

Example 4: A man 1.5m tall standing at a distance of 50m from a tree observes the angle of elevation of the top of the tree is 45°. Find the height of the tree.

Solution:

Let CE be the height of the tree, AB be the height of the man and BC be the distance between the foot of the tree and the man.

Here, AB = 1.5m, BC = 50m, Angle of elevation DAE = 45°

In rectangle ABCD,

DC = AB = 1.5m and AD = BC = 50m

tan45° = p/b = ED/AD = ED/50

or,     1 = ED/50

or,     ED = 50m

Now, CE = ED + DC = 50 + 1.5 = 51.5m

So, the required height of the tree is 51.5m.

Example 5: A boy, 1.2m tall, is flying a kite. When the length of the string of the kite is 180m, it makes an angle of 30° with the horizontal line. At what height is the kite from the ground?

Solution:

Let CK be the height of the kite from the ground, AB be the height of the boy and AK be the length of the string.

Here, AB = 1.2m, AK = 180m, Angle of elevation DAK = 30°

In rectangle ABCD, AB = DC = 1.2m.

sin30° = p/h = KD/AK = KD/180

or,     ½ = KD/180

or,     2KD = 180

or,     KD = 180/2 = 90m

Now, CK = KD + DC = 90 + 1.2 = 91.2m

So, the kite is at a height of 91.2m from the ground.

Example 6: A man is 1.6m tall and the length of his shadow in the sun is 1.6√3 m. Find the altitude of the sun.

Solution:

Let AB be the height of the man and BC be the length of his shadow. Let θ be the altitude of the sun.

Here, AB = 1.6 m, BC = 1.6√3 m.

In right angled triangle ABC,

tanθ = p/b = AB/BC

or,     tanθ = 1.6/1.6√3

or,     tanθ = 1/√3 = tan30°

θ = 30°

So, the required altitude of the sun is 30°.

Example 7: A man √3 m tall is 72 m away from a tower 25√3 m high. Find the angle of elevation of the top of the tower from his eyes.

Solution:

Let AB be the height of the man, CD be the height of the tower and BD be the distance between the man and the tower.

Here, BD = AE = 72 m

AB = √3 m

CD = 25√3 m

CE = CD – ED = 25√3 – AB = 25√3 - √3 = 24√3 m.

Now, in right angled triangle AEC,

tanA = p/b = CE/AE

or,     tanA = 24√3/72 = √3/3

or,     tanA = 1/√3

or,     tanA = tan30°

A = 30°

So, the required angle of elevation is 30°.

Example 8: The top of a tree which is broken by the wind makes an angle of 60° with the ground at a distance 3√3 m from the foot of the tree. Find the height of the tree before it was broken.

Solution:

Let AC be the height of the tree before it was broken, BD be the broken part of the tree and CD be the distance between the foot of the tree and the point on the ground at which the top of the tree touched.

Here, CD = 3√3 m, Angle of elevation, CDB = 60°.

In right angled triangle BCD,

tan60° = p/b = BC/CD = BC/3√3

or,     √3 = BC/3√3

or,     BC = 9 m

Also, cos60° = b/h = CD/BD

or,     ½ = 3√3/BD

or,     BD = 6√3 = 6 × 1.732 = 10.39 m.

Then, the height of the tree before broken = AC

= AB + BC

= BD + 9m

= 10.39m + 9m

= 19.39m

So, the required height of the tree was 19.39 m.

Example 9: Two vertical poles are fixed 60 m apart. The angle of depression of the top of the first pole from the top of the second pole which is 150 m high is 30°. Find the height of the first pole.

Solution:

Let AB be the height of the first (shorter) pole, CD be the height of the second (taller) pole and BD be the distance between them.

Here, BD = AE = 60 m

CD = 150 m

The angle of depression = angle of elevation, CAE = 30°.

In right angled triangle CEA,

tan30° = p/b = CE/AE = CE/60

or,     1/√3 = CE/60

or,     CE = 60/√3

or,     CE = 34.64 m

Again, in rectangle ABDE,

AB = DE = CD – CE = 150 – 34.64 = 115.36 m

So, the required height of the first pole is 115.36 m.

Example 10: Two men are on the opposite side of a tower of 30 m high. They observed the angle of elevation of the top of the tower and found to be 30° and 60° respectively. Find the distance between them.

Solution:

Let AB be the height of the tower and CD be the distance between the two men.

Here, AB = 30m.

Angle of elevations, BCA =30° and BDA = 60°.

In right angled triangle ABC,

tan30° = p/b = AB/BC

or,     1/√3 = 30/BC

or,     BC = 30√3 = 30 × 1.732 = 52.05 m

Again, in right angled triangle ABD,

tan60° = p/b = AB/BD

or,     √3 = 30/BD

or,     BD = 30/√3 = 30/1.732 = 17.32m

Now, CD = BC + BD = 52.05 + 17.32 = 69.37 m

So, the required distance between the men is 69.37m.

You can comment your questions regarding the problems on height and distance here.