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