Question

For a man, the angle of elevation of the highest point of the temple situated east of him is $60 ^ { \circ }$. On walking 240 metresto north, the angle of elevation is reduced to $30 ^ { \circ }$ then the height of the temple is

• A. $60 \sqrt { 6 } m$
• B.
• C. $50 \sqrt { 3 } \mathrm {~m}$
• D.

Answer 1

Answer: (A)

$60 \sqrt { 6 } m$

Answer 2

Total distance from temple =$\sqrt { x ^ { 2 } + ( 240 ) ^ { 2 } }$

where $x = \frac { h } { \tan 60 ^ { \circ } } = \frac { h } { \sqrt { 3 } }$

So distance = $\sqrt { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } }$, but $\frac { h } { \sqrt { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } } } = \frac { 1 } { \sqrt { 3 } }$

⇒ $\frac { h ^ { 2 } } { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } } = \frac { 1 } { 3 }$

After solving $h = 60 \sqrt { 6 }$ metre

Total distance from temple =$\sqrt { x ^ { 2 } + ( 240 ) ^ { 2 } }$

where $x = \frac { h } { \tan 60 ^ { \circ } } = \frac { h } { \sqrt { 3 } }$

So distance = $\sqrt { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } }$, but $\frac { h } { \sqrt { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } } } = \frac { 1 } { \sqrt { 3 } }$

⇒ $\frac { h ^ { 2 } } { \frac { h ^ { 2 } } { 3 } + ( 240 ) ^ { 2 } } = \frac { 1 } { 3 }$

After solving $h = 60 \sqrt { 6 }$ metre