sir this is the important question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
From a tower 126 m high,
the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16 and 12 20’.
Find the distance between the rocks if they are on
(i) the same side of the tower
(ii) the opposite sides of the tower.
question no 30 , heights and distances , ICSE board
Consider CD as the tower of height = 126 m
A and B are the two rocks on the same line
Angles of depression are 160 and 120 20’
In triangle CAD
tan θ = CD/AD
Substituting the values
tan 160 = 126/x
0.2867 = 126/x
So we get
x = 126/0.2867
x = 439.48
In right triangle CBD
tan 120 20’ = 126/y
So we get
0.2186 = 126/y
y = 126/0.2186 = 576.40
(i) In the first case
On the same side of the tower
AB = BD – AD
AB = y – x
Substituting the values
AB = 576.40 – 439.48
AB = 136.92 m
(ii) In the second case
On the opposite side of the tower
AB = BD + AD
AB = y + x
Substituting the values
AB = 576.40 + 439.48
AB = 1015.88m