Question

The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60∘ and 30∘ respectively. Then height of the temple is ?

​

Answer 1

Given :-

Height of building = 30m

The angle of elevation of the top of the temple from the foot = 60°

The angle of elevation from top of the building = 30°

Solution :-

In ΔDBC ,

Tan 60° = p / B

Put the values,

Tan 60° = x + 30m / BC

√3 = x + 30 / BC

BC = x + 30 / √3

In ΔDAE ,

As we know AE = BC

DE / AE = tan 30°

put the value,

x/ x + 30 / √3 = 1 / √3

√3x = x + 30 / √3

3x = x + 30

3x - x = 30

2x = 30

x = 30/2

x = 15

Now, h = x + 30m

put the values ,

H = 15 + 30 = 45m

Height of the Temple is 45m

{ Note : - Refer the above attachment }

Answer 2

Step-by-step explanation:

In △DBC,

tan60∘ = x+30/BC

BC= x + 30 / √3 ....(1)

In △DAE,

DE/AE =tan30∘

Now, AE=BC = x / x + 30 / √3 = 1 / √3 = √3x = x + 30/√3

3x=x+30

2x=30⇒x=15 m

h=x+30=15+30=45 m.

Hope this is helpful for you.