sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

it is given that In the figure, not drawn to scale, TF is a tower.

The elevation of T from A is x0 where tan x = 2/5 and AF = 200 m.

The elevation of T from B, where AB = 80 m, is y0.

Calculate: (i) the height of the tower TF.

(ii) the angle y, correct to the nearest degree.

question no 22 , heights and distances , ICSE board

Consider the height of the tower TF = x

It is given that

tan x = 2/5, AF = 200 m, AB = 80 m

(i) In right triangle ATF

tan x

^{0}= TF/AFSubstituting the values

2/5 = x/200

So we get

x = (2 × 200)/ 5

x = 400/5

x = 80 m

Hence, the height of tower is 80 m.

(ii) In right triangle TBF

tan y = TF/BF

Substituting the values

tan y = 80/ (200 – 80)

tan y = 80/120

tan y = 2/3 = 0.6667

So we get

y = 33

^{0}41’ = 34^{0}