## Solution of NCERT Exercise 9.1

Question: 1 - A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).

Answer: In this figure; AC is the rope which is 20 m long and we have to find AB.

In Δ ABC;

Hence; the height of the pole is 10 m

In Δ ABC;

Hence; the height of the pole is 10 m

Question: 2 - A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer: Let BD is the tree. Because of storm AD is broken and bends as AC.

Here, AC is the broken part of the tree and AB is the part which is still upright.

Here, BC = 8 m, and ∠ BCA=30

Sum of AB and AC, which is equal to BD shall give the height of the tree.

In Δ ABC;

Now;

Thus, height of the three was 8√3 m

Here, AC is the broken part of the tree and AB is the part which is still upright.

Here, BC = 8 m, and ∠ BCA=30

^{0}Sum of AB and AC, which is equal to BD shall give the height of the tree.

In Δ ABC;

Now;

Thus, height of the three was 8√3 m

Question: 3 - A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer: In the first case; p = 1.5 m and angle of elevation = 30°

We need to find the hypotenuse.

In the second case; p = 3 m, angle of elevation = 60° and h = ?

Thus, lenght of slide in first case is 3m and in second case is 2√3 m

We need to find the hypotenuse.

In the second case; p = 3 m, angle of elevation = 60° and h = ?

Thus, lenght of slide in first case is 3m and in second case is 2√3 m

Question: 4 - The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer: Here; b = 30 m, angle of elevation = 30° and p = ?

Thus, height of the tower is 10√3 m

Thus, height of the tower is 10√3 m

Question: 5 - A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer: Here; p = 60 m, angle of elevation = 60° and h = ?

Question: 6 - A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer: Here; p = 30 – 1.5 = 28.5 m, angle of elevation in first case = 30° and in second case = 60°. Difference between bases in the first case and second case will give the distance covered by the boy.

1st case:

Distance covered by the boy can be calculated as follows:

1st case:

Distance covered by the boy can be calculated as follows:

Question: 7 - From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer: Let us assume the height of building = p = 20 m

Total height of building and transmission tower = p’

The angle of elevation to the top of the building = 45° and that to the top of transmission tower = 60°

Height of building subtracted from total height will give the height of transmission tower.

1st case:

2nd case:

Height of transmission tower can be calculated as follows:

Total height of building and transmission tower = p’

The angle of elevation to the top of the building = 45° and that to the top of transmission tower = 60°

Height of building subtracted from total height will give the height of transmission tower.

1st case:

2nd case:

Height of transmission tower can be calculated as follows:

Question: 8 - A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Answer: In this figure; AD is the height of the statue which is 1.6 m, DB is the height of the pedestal and C is the point where observer is present.

In Δ ABC;

In Δ DBC;

From equation (1) and (2);

In Δ ABC;

In Δ DBC;

From equation (1) and (2);

Question: 9 - The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer: In this figure; DC = 50 m, ACB = 30° and DBC = 60°; we have to find AB.

In Δ DCB;

In Δ ABC;

In Δ DCB;

In Δ ABC;

Question: 10 - Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer: In this figure; BD = 80 m, ACB = 30°, ECD = 60° and AB = ED = ?

In Δ ABC;

In Δ EDC;

So, from equation (1) and (2);

By substituting the value of BC in equation (1) we get;

Moreover, BC = 60 m and CD = 20 m

In Δ ABC;

In Δ EDC;

So, from equation (1) and (2);

By substituting the value of BC in equation (1) we get;

Moreover, BC = 60 m and CD = 20 m

Question: 11 - A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Answer: In this figure; CD = 20 m, ACB = 60°, ADB = 30°, AB = ? and BD = ?

In Δ ABC;

In Δ ABD;

From equation (1) and (2);

Substituting the value of BC in equation (1) we get;

In Δ ABC;

In Δ ABD;

From equation (1) and (2);

Substituting the value of BC in equation (1) we get;

Question: 12 - From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Answer: In this figure; AB = 7m, ACB = 45°, EAD = 60° AB = DC and EC = ED + DC ?

In Δ ABC;

In Δ EDA; AD = BC = 7 m

Height of tower can be calculated as follows:

In Δ ABC;

In Δ EDA; AD = BC = 7 m

Height of tower can be calculated as follows:

Question: 13 - As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer: In this figure; AB = 75 m, ACB = 45°, ADB = 30°, BC = ? CD = ?

In Δ ABC;

In Δ ABD;

In Δ ABC;

In Δ ABD;

Question: 14 - A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Answer: In this figure; BF = 1.2 m AG = 88.2 m, ABC = 60°, EBD = 30° and we have to find CD

AC = AG – GC = 88.2 – 1.2 = 87 m

In Δ ABC;

In Δ EBD;

Distance covered by balloon;

AC = AG – GC = 88.2 – 1.2 = 87 m

In Δ ABC;

In Δ EBD;

Distance covered by balloon;

Question: 15 - A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer: In this figure; ADB = 30°, ACB = 60° and time taken to reach from D to C = 6 seconds We need to find the time taken to reach from C to B

In Δ ABD;

In Δ ABC;

From equation (1) and (2);

Time taken to reach from C to B = 6 second

In Δ ABD;

In Δ ABC;

From equation (1) and (2);

Time taken to reach from C to B = 6 second

Question: 16 - The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer: In this figure; ADB = θ and ACB = 90°-θ

In Δ ABD;

In Δ ABD;

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