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 x0 = TF/AF
Substituting 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 = 330 41’ = 340