Question

The angular elevation of a tower CD at a point A due south of it is 60^o and at a point B due west of A, the elevation is $A B = 3$km, the height of the tower is

• A. $2 \sqrt { 3 }$km
• B. $2 \sqrt { 6 }$ km
• C. $\frac { 3 \sqrt { 3 } } { 2 }$ km
• D. $\frac { 3 \sqrt { 6 } } { 4 }$ km

Answer 1

Answer: (D)

$\frac { 3 \sqrt { 6 } } { 4 }$ km

Answer 2

In $\triangle C B D$, $\tan 30 ^ { \circ } = \frac { h } { B C } \Rightarrow B C = \sqrt { 3 } h$ ,

In $\triangle A C D$, $\tan 60 ^ { \circ } = \frac { h } { A C } \Rightarrow A C = \frac { h } { \sqrt { 3 } }$

Now, $27 = 10 h ^ { 2 }$

$h ^ { 2 } = \frac { 27 } { 10 } \Rightarrow h = \frac { 3 \sqrt { 3 } } { \sqrt { 10 } } = \frac { 3 \sqrt { 6 } } { \sqrt { 20 } } \simeq \frac { 3 \sqrt { 6 } } { 4 }$ km