# 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.

**Answer:** 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.

**Answer:** 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.

**Answer:** 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.

**Answer:** 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.

**Answer:** 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.

**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;

`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.

**Answer:** 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