Height and Distance
NCERT Exercise 9.1
Part 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°.
Answer: In this figure; AC is the rope which is 20 m long and we have to find AB.
In Δ ABC;
`text(sin)θ=p/h=(AB)/(AC)`
Or, `text(sin) 30°=(AB)/(AC)`
Or, `(1)/(2)=(AB)/(20)`
Or, `AB=10 cm`
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=30o
Sum of AB and AC, which is equal to BD shall give the height of the tree.
In Δ ABC;
`text(cos)θ=b/h=(BC)/(AC)`
Or, `text(cos) 30°=(8)/(AC)`
Or, `(sqrt3)/(2)=(8)/(AC)`
Or, `AC=(16)(sqrt3)`
Now;
`text(tan)θ=p/b=(AB)/(BC)`
Or, `text(tan) 30°=(AB)/(8)`
Or, `(1)/(sqrt3)=(AB)/(8)`
Or, `AB=(8)/(sqrt3)`
Height of tree = AB + AC
`(8)/(sqrt3)+(16)/(sqrt3)`
`=(24)/(sqrt3)=(24)/(sqrt3)xx(sqrt3)/(sqrt3)=8sqrt3 m`
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.
`text(sin)θ=p/h`
Or, `text(sin) 30°=(1.5)/(h)`
Or, `1/2=(1.5)/(h)`
Or, `h=3 m`
In the second case; p = 3 m, angle of elevation = 60° and h = ?
`text(sin)θ=p/h`
Or, `text(sin) 60°=3/h`
Or, `(sqrt3)/(2)=3/h`
Or, `h=(6)/(sqrt3)=2sqrt3`
Thus, length 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 = ?
`text(tan) θ=p/b`
Or, `text(tan) 30°=(p)/(30)`
Or, `(1)/(sqrt3)=(p)/(30)`
Or, `p=(30)/(sqrt3)`
`=(30)/(sqrt3)xx(sqrt3)/(sqrt3)=10sqrt3`
Thus, height of the tower is `10sqrt3` 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 = ?
`text(sin) θ=p/h`
Or, `text(sin) 60°=(60)/(h)`
Or, `1/2=(60)/(h)`
Or, `h=(120)/(sqrt3)`
`=(120)/(sqrt3)xx(sqrt3)/(sqrt3)=40sqrt3 m`
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:
`text(tan) 30°=p/b`
Or, `(1)/(sqrt3)=(28.5)/(b)`
Or, `b=28.5sqrt3`
2nd case:
`text(tan) 60°=p/b`
Or, `sqrt3=(28.5)/(b)`
Or, `b=(28.5)/(sqrt3)`
Distance covered by the boy can be calculated as follows:
`28.5sqrt3-(28.5sqrt3)/(sqrt3)`
`=(28.5xx3-28.5)/(sqrt3)`
`=(57)/(sqrt3)=19sqrt3 m`
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:
`text(tan) 45°=p/b`
Or, `(20)/(b)=1`
Or, `b=20 m`
2nd case:
`text(tan) 60°=(p’)/(b)`
Or, `sqrt3=(p’)/(20)`
Or, `p’=20sqrt3`
Height of transmission tower can be calculated as follows:
`p'- p = 20sqrt3 - 20`
`= 20(sqrt3 - 1)`
`= 20(1.732 - 1)`
`= 20 x× 0.732 = 14.64 m`
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;
`text(tan) 60°=(p)/(b)`
Or, `sqrt3=(DB+1.6)/(b)`
Or, `b=(DB+1.6)/(sqrt3)` --------(1)
In Δ DBC;
`text(tan) 45°=p/b`
Or, `1=(DB)/(b)`
Or, `b=DB` ----------(2)
From equation (1) and (2);
`DB=(DB+1.6)/(sqrt3)`
Or, `DBsqrt3=DB+1.6`
Or, `Dbsqrt3-DB=1.6`
Or, `DB(sqrt3-1)=1.6`
Or, `DB=(1.6)(sqrt3-1)`
`=(1.6)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)`
`=(1.6)/(3-1)xx(sqrt3+1)`
`=0.8(sqrt3+1) m`
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;
`text(tan) 60°=p/b`
Or, `sqrt3=(50)/(b)`
Or, `b=(50)/(sqrt3)`
In Δ ABC;
`text(tan) 30°=p/b`
Or, `(1)/(sqrt3)=(AB)/((50)/(sqrt3))`
Or, `AB=(50)/(sqrt3xxsqrt3)=16(2)/(3) m`