Frequently Asked Questions on Chapter 9 Some Applications of Trigonometry
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.
Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.
To Find: AB (height of the tower)
Let us consider in △ABC
tan 30° = AB/BC
1/√3 = AB/30
AB = 10√3
Hence, the height of the tower is 10√3 m.
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.
Let us consider in △ABC
Let BC = Height of the kite from the ground, BC = 60 m
AC = Inclined length of the string from the ground and A is the point where the string of the kite is tied.
To Find: Length of the string from the ground i.e. the value of AC
So,
sin 60° = BC/AC
√3/2 = 60/AC
AC = 40√3 m
Hence, the length of the string from the ground is 40√3 m.
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.
Let BC be the 20m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower = AB = AC – BC
To Find: AB, Height of the tower
Let us consider in right ΔBCD,
tan 45° = BC/CD
1 = 20/CD
CD = 20
Again,
In right ΔACD,
tan 60° = AC/CD
√3 = AC/20
AC = 20√3
Now, AB = AC – BC = (20√3 – 20) = 20(√3 – 1)
Hence, the height of the transmission tower is 20(√3 – 1) m.
In right ΔABC,
tan x = AB/BC
tan x = AB/4
AB = 4 tan x … (i)
Again, from right ΔABD,
tan (90°-x) = AB/BD
cot x = AB/9
AB = 9 cot x … (ii)
Multiplying equation (i) and (ii)
AB2 = 9 cot x × 4 tan x
⇒ AB2 = 36 (because cot x = 1/tan x
⇒ AB = ± 6
Since height cannot be negative. Therefore, the height of the tower is 6 m.
Hence Proved.