Practice Set For Height & Distance

Earlier on the topic ‘Height & Distance’ we have discussed the basic definition of angle of elevation, depression, and some basic straight forward questions. And from here we will discuss in more detail and learn the new type of question which usually comes into the exam and will also try to solve those questions in a tricky way.

Some Basic Concepts:-

Angle of Elevation:- The angle from a point on the ground to the point above its height is called the angle of elevation. In

other words when we see above then the angle on the eye to the horizontal is called the angle of elevation.

Here in the given image, a person is looking at a point a which is above the horizontal line of his eyesight. So here angle made on the eye i,e; ∠A is called the angle of elevation.

Angle of Depression:- The angle from a point above the ground to the point on the ground is called the angle of depression. In other words, when we see below then the angle with eye to the horizontal is called the angle of depression.

Here in the given image, a person is looking at a point that is below the horizontal line of his eyesight. So here angle made on the eye i,e; ∠a is called angle of depression.

Some tricks that will help in solving questions quickly:-

In maximum questions, angle given are mostly 30°. 45°, and 60°, so will use the ratio of these angles to solve the question.

CASE – 1, When the Angle is 30° :-

We know that sin 30° = p/h = 1/2

So, b = √(4 – 1) = √3

Then, if an angle of elevation/depression is given as 30°, then the ratio of p, b, h is 1 : √3 : 2

Question-based on Case – 1.

Question: 1 If a man is 20 m apart from a pole and the angle of elevation from the top of the pole is 30°, then find the height of the pole?

Solution:

Here BC = 20m

According to given trick, p, b, h is 1 : √3 : 2

Here b = 20m

And, p/b = AB/BC = 1/√3

⇒ AB/20 = 1/√3

So, AB = 20/√3 m

So, the required value is 20/√3.

CASE – 2, When angle is 45° :-

We know that tan 45° = p/b = 1/1

So, h = √(1 + 1) = √2

Then, if an angle of elevation/depression is given as 45°, then the ratio of p, b, h is 1 : 1 : √2

Question-based on Case – 2.

Question: 2 If the length of a pole is 12m and the angle of elevation from the top of the pole to point A on the ground is 45°, then find the distance of the pole from point A?

Solution:

Here AB = 12m

According to given trick, p, b, h is 1 : 1 : √2

Here p = 12m

And, p/b = BC/AC = 1/1

⇒ 12/AC = 1/1

So, AC = 12 m

So, the required value is 12m.

CASE – 3, When angle is 60° :-

We know that tan 60° = p/b = √3/1

So, h = √(3 + 1) = 2

Then, if an angle of elevation/depression is given as 60°, then the ratio of p, b, h is √3 : 1 : 2

Now let’s practice some of the questions based on the latest exams:

Question: 3 A pole is broken at some height and makes  60° to the ground and the distance between the bottom point of the pole and the point it touches the ground is 10m. What is the length of the pole?

Let the pole is broken from point B and it touches point A on the ground.

Since angle of elevation is 60° so  p, b, h is √3 : 1 : 2

Then, tan 60° = p/b = √3/1

⇒ BC/AC = √3/1

⇒ BC/10 = √3/1

⇒ BC = 10√3 m = 17.3

And, AB = √(BC2 + AC2) = √(300 + 100)m = 20m (Approx)

Since the pole is broken from the point B so the length of pole = BC + AB = (17.3 + 20)m = 37.3m = 37 m (approximately)

The required length is 37m.

Question: 4 The shadow of a tree decreases by 15m when the sun’s altitude changes from 45° to 60°. Find the length of the tree.

Solution:

Let AB be the length of the tree and the length of shadow be x when the angle of elevation is 45°.

Then, the length of shadow when the angle of elevation is 60° = x – 15m

For angle 45° in triangle ABD, we have

AB/BD = 1/1 = AB/x

So, AB = x

Now, for angle 60° in triangle ABC, we have

AB/BC = √3/1 = x/(x – 15)

⇒ x = √3x – 15√3

⇒ (√3 – 1)x = 15√3

⇒ x = 15√3/(√3 – 1)m

The length of the tree is 15√3/(√3 – 1)m.

Question: 5 A pole stands upright on the ground with the help of two wires to its top and affix to the ground on opposite sides. If the angle of elevation for both wires are 30° and 60° respectively and the length of the first wire is 8m, then find the length of another wire.

Solution:

Let AD be the length of the pole and AB and AC be the length of the first and second wires respectively.

In triangle ABD, we have

∠B = 30°

So, AB : AD = h : p = 2 : 1

⇒ 8/AD = 2/1

⇒ AD = 4m

Now, in triangle ADC, we have

∠C = 60°

So, AD : AC = √3/2

⇒ 4/AC = √3/2

⇒ AC = 8/√3 = 8√3/3 m

The required length of wire is 8√3/3 m.

Question: 6 Angle of depression from a kite flying into the sky to a point on the ground is 45° and it changes to 60° when the kite flies 10m higher. Find the initial height of the kite.

Solution:

Let the kite was flying at a height of AB initially.

According to the question, in triangle ABC, we have

∠C = 45°

So, AB = BC = 1/1

Then, AB = BC

Now, in triangle ADC, we have

∠C = 60°

So, AD/BC = √3/1

⇒ (AB + 10)/AB = √3

⇒ AB + 10 = AB√3

⇒ AB(√3 – 1) = 10m

⇒ AB = 10/(√3 – 1) = 5(√3 + 1)m

The required length is 5(√3 + 1)m.

Question: 7 The angle of elevation from the top of building A to the top and bottom of building B is 30° and 60° respectively. If these buildings are 3√3m apart, then find the length of the higher building.

Solution:

Let AB be the length of building A and CD be the length of building B.

According to the figure BC = AE = 3√3m. and AB = EC

Now, in triangle ADE, we have

∠A = 30°

So, DE : AE = p : b = 1 : √3

⇒ DE/3√3 = 1/√3

⇒ DE = 3m

Now, in triangle ABC, we have

∠C = 60°

So, AB : BC = p : b = √3 : 1

⇒ AB/3√3 = √3/1

⇒ AB = 9m

So, length of building B = CE + DE = 9m + 3m = 12m

The required length is 12m.

Question: 8 A man is travelling in the east direction from his home to his office. When he traveled 15m altitude from the roof of his house is 60° and when he walks furthermore 5 seconds the altitude becomes 30°. What is his speed?

Solution:

Let AB be the height of his house and initially he covers AD distance and after 5 seconds he reaches point C.

In triangle ABD, we have

AD = 15m

∠D = 60°

So, AB : AD = √3/1

⇒ AB/15 = √3

⇒ AB = 15√3 m

Now, in triangle ABC, we have

∠C = 30°

So, AB/AC = 1/√3

⇒ 15√3/AC = 1/√3

⇒ AC = 45m

So, CD = AC – AD = 45m – 15m = 30m

Since CD is covered in 5 seconds,

So required speed = 30/5 = 6m/s

The required speed is 6m/s.

Question: 9 There are two poles across the road. If the width of the road is ‘a’ and from the middle point of the road, the angle of elevation to both poles is 45° and 30° respectively, then find the difference between the height of the poles.

Solution:

Let AB and CD be two poles and O be the midpoint of the road.

In triangle OCD, we have

∠COD = 45°

So, CD/OD = 1/1

⇒ CD = a/2

Now, in triangle AOB, we have

∠AOB = 30°

So, AB/BO = 1/√3

⇒ AB/(a/2) = 1/√3

⇒ AB = a/2√3

So, the required difference = a/2 – a/2√3 = (a/2)(1 – 1/√3) = (a/2)(1.732 – 1)/√3 = (0.732/√3)a

The required difference is  (0.732/√3)a.

Question: 10 In the given figure if BC = 1m, then find the value of AC.

Solution:

In the given triangle, we have

BC = 1m and ∠C = 30°

Then, BC : AC = b : h = √3 : 2

⇒ 1/AC = √3/2

⇒ AC = 2/√3

The required value is 2/√3.