# Height and Distance

## NCERT Exercise 9.1

### Part 2

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.

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

In Δ ABC;

text(tan) 30°=p/b

Or, (1)/(sqrt3)=(AB)/(BC)

Or, AB=(BC)/(sqrt3) -------- (1)

text(tan) 60°=p/b=(EC)/(CD)

Or, sqrt3=(ED)/(80-BC)

Or, ED=sqrt3(80-BC) --------- (2)

Since AB = ED

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

In Δ EDC;
(BC)/(sqrt3)=sqrt3(80-BC)

Or, BC=3(80-BC)
Or, BC=240-3BC
Or, 4BC=240
Or, BC=60 m

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

AB=(60)/(sqrt3)=20sqrt3 m

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.

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

In Δ ABC; text(tan) 60°=p/b

Or, sqrt3=(AB)/(BC)
Or, AB=BCsqrt3 --------- (1)

In Δ ABD; text(tan) 30°=p/b

Or, (1)/(sqrt3)=(AB)/(BC+20)

Or, AB=(BC+20)/(sqrt3) ---------- (2)

From equation (1) and (2);
BC\sqrt3=(BC+20)/(sqrt3)

Or, 3BC=BC+20
Or, 2BC=20
Or, BC=10 m

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

AB = 10sqrt3 m which is the height of the tower

Width of canal = 10 m

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.

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

In Δ ABC;
text(tan) 45°=p/b

Or, 1=(AB)/(BC)

Or, BC=AB=7 m

In Δ EDA; AD = BC = 7  m
text(tan) 60°=p/b

Or, sqrt3=(ED)/(7)

Or, ED=7sqrt3 m

Height of tower can be calculated as follows:

EC = ED + DC = 7sqrt3 + 7 = 7(sqrt3 +1)  m

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.

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

In Δ ABC;
text(tan) 45°=p/b

Or, 1=(75)/(BC)

Or, BC=75 m

In Δ ABD;
text(tan) 30°=p/b

Or, (1)/(sqrt3)=(75)/(BD)

Or, BD=75sqrt3 m
Or, BC+CD=75sqrt3
Or, 75+CD=75sqrt3
Or, CD=75sqrt3-75
=75(sqrt3-1) m

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.

Solution: In this figure; BF = 1.2 m AG = 88.2 m, AC = AG – GC = 88.2 – 1.2 = 87 m ∠EBD = 30° and ∠ABC = 60°, we have to find CD

In Δ ABC;
text(tan) 60°=p/b
Or, sqrt3=(87)/(BC)
Or, BC=(87)/(sqrt3

In Δ EBD;
text(tan) 30°=p/b
Or, (1)/(sqrt3)=(87)/(BD)
Or, BD=87sqrt3

Distance covered by balloon;
CD=BD-BC
Or, CD=87sqrt3-(87)/(sqrt3)
=(87xx3-87)/(sqrt3)=(174)/(sqrt3)
=58sqrt3 m

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.

Solution: 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;
text(tan) 30°=p/b
Or, (1)/(sqrt3)=(AB)/(BD)
Or, AB=(BD)/(sqrt3) --------- (1)

In Δ ABC;
text(tan) 60°=p/b
Or, sqrt3=(AB)/(BC)
Or, AB=BCsqrt3 ----------- (2)

From equation (1) and (2);
(BD)/(sqrt3)=BCsqrt3
Or, (BC+6)/(sqrt3)=BCsqrt3
Or, 3BC=BC+6
Or, 2BC=6
Or, BC=3

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.

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

In Δ ABD;
text(tan) θ=(AB)/(DB)
Or, text(tan) θ=(AB)/(9) -------(1)

In Δ ABC;
text(tan)(90°-θ)=(AB)/(BC)
Or, text(cot) θ=(AB)/(4) ---------(2)

We know;
text(cot) θ=(1)/(text(tan) θ)

Hence;
(AB)/(4)=(9)/(AB)
Or, (AB)^2=9xx4=36
Or, AB=sqrt(36)=6 m proved