NCERT Solutions For Class 10 Maths Chapter 9 Some Application of Trigonometry
Exercise 9.1
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:
BC = `x`
CD = `80 - x`
In `triangleABC`
`tan 60° = frac{AB}{BC}`
`⇒sqrt3 = frac{AB}{x}`
`⇒ AB = sqrt3x------(i)`
In `triangleEDC`
`tan 30° = frac{ED}{CD}`
`⇒1/sqrt3 = frac{ED}{80 -x}`
`⇒ ED = frac{80 - x}{sqrt3}-------(ii)`
From equation (i) & (ii)
`sqrt3x = frac{80 - x}{sqrt3}`
`⇒ 3x = 80 - x`
`⇒ 4x = 80`
`⇒ x = 80/4`
`⇒ x = 20`
`therefore` BC = 20m
`therefore CD = 80 - x= 80 -20` = 60 m
In `triangleABC`
`tan 60° = frac{AB}{BC}`
`⇒sqrt3 = frac{AB}{20}`
`therefore AB = 20sqrt3 m`
Height of the poles = `20sqrt3 m`
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° (see Fig.). Find the height of the tower and the width of the canal.
Solution:
Let the width, BC of the canal be `x`.
Let the height of the TV tower be `h`
In `triangleABC`
`tan 60° = frac{AB}{BC}`
`⇒ sqrt 3 = h/x`
`⇒ h = sqrt 3x ----(i)`
In `triangleABD`
`tan 30° = frac{AB}{BD}`
`⇒ 1/sqrt 3 = frac{h}{20 + x}`
`⇒ h = frac{20 + x}{sqrt3} ----(ii)`
from equation (i) & (ii)
`sqrt 3x = frac{20 + x}{sqrt3}`
`⇒ 3x = 20 +x`
`⇒ 2x = 20`
`x = 10`
`therefore` width of the canal is 10 m
In `triangleABC`
`tan 60° = frac{AB}{BC}`
`⇒ sqrt 3 = h/10`
`⇒ h = 10sqrt 3`
`therefore` height of the TV tower is `10sqrt3` m
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:
ED = `x`
AD = BC
AB = CD = 7 m
In `triangleABC`
tan45° = AB/BC
`1 = frac{7}{BC}`
`therefore` BC = 7m
In `triangleADE`
tan60° = ED/AD
`sqrt3 = x/7`
`x = 7sqrt3`
`therefore` height of the tower EC = ED + CD
`= 7sqrt3 + 7`
`= 7(sqrt3 + 1)m`
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 `triangleABC`
tan 60° = `frac{AB}{BC}`
`⇒ sqrt3 = frac{AB}{BC}`
`therefore AB = sqrt3BC------(i)`
In `triangleABD`
`tan 30° = frac{AB}{BD}`
`⇒ sqrt3AB = BD`
`therefore AB = frac{BD}{sqrt3}-----(ii)`
from equation (i) & (ii)
`sqrt3BC = frac{BD}{sqrt3}`
`⇒ 3BC = BC + CD`
`⇒ 2BC = CD`
`therefore BC = frac{CD}{2}`
The distance BC is half of CD
`therefore` the time taken is also half.
The car taken 6sec to travel CD, so the time taken by the car to travel BC = 6/2 = 3sec.
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 `triangleABC`
`tan x = frac{AB}{BC}`
`⇒ tan x = frac{AB}{4}`
`⇒ AB = 4tan x----(i)`
In `triangleABD`
`tan (90° - x) = frac{AB}{BD}`
`⇒ tan (90° - x) = frac{AB}{9}`
`⇒ AB = 9tan (90° - x) ----(ii)`
Multiply equations (i) & (ii)
`AB^2 = 4 tan x × 9 cotx`
`AB^2 = 36 tanx × 1/tan x`
`AB = sqrt36`
`therefore AB = 6 `
Height of the tower is 6m.
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