Question

There are two temples, one on each bank of a river, just opposite to each other. One temple is 50 m high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the width of the river and the height of the other temple.

Answer 1

The width of river is 28.86 meters

The height of other temples is 66.67 meters

Step-by-step explanation:

Given as :

The height of temple 1 = H = 50 meters

The height of temple 2 = h meters

The angle of depression of the top of a temple = 30°

The angle of depression of its foot =  60°

The width of river = distance between two temples  = x meters

According to question

From figure

In Δ AEB

Tan angle = $\dfrac{perpendicular}{base}$

Tan  30° = $\dfrac{AB}{BE}$

$\dfrac{1}{\sqrt{3} }$   = $\dfrac{h-50}{x}$

Or,  x = √3 (h - 50)            ...........1

Again

In Δ ECD

Tan angle = $\dfrac{perpendicular}{base}$

Tan  60° =  $\dfrac{ED}{CD}$

√3 = $\dfrac{50}{x}$

Or,  x =  $\dfrac{50}{\sqrt{3} }$                  ..............2

So, The width of river = x = 28.86   meters

Solving eq 1 and eq 2

$\dfrac{50}{\sqrt{3} }$  =  √3 (h - 50 )

Or, √3 h  = $\dfrac{50}{\sqrt{3} }$ + 50 √3

Or, h = $\dfrac{50}{3}$ + 50

∴   h = 16.67 + 50 = 66.67 meters

So, height of other temple = 66.67 meters

Hence, The width of river is 28.86 meters

The height of other temples is 66.67 meters    Answer