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Question: A tower subtends angles \(\alpha , 2 \alpha , 3 \alpha\)respectively at points A, B and C, all lying...

A tower subtends angles α,2α,3α\alpha , 2 \alpha , 3 \alpharespectively at points A, B and C, all lying on a horizontal line through the foot of the tower. Then ABBC=\frac { A B } { B C } =

A

sin⁡3αsin⁡2α\frac { \sin 3 \alpha } { \sin 2 \alpha }

B

1+2cos⁡2α1 + 2 \cos 2 \alpha

C

2+cos⁡3α2 + \cos 3 \alpha

D

sin⁡2αsin⁡α\frac { \sin 2 \alpha } { \sin \alpha }

Answer

1+2cos⁡2α1 + 2 \cos 2 \alpha

Explanation

Solution

From sine rule

⇒ BEsin⁡(180∘−3α)=BCsin⁡α\frac { B E } { \sin \left( 180 ^ { \circ } - 3 \alpha \right) } = \frac { B C } { \sin \alpha }

⇒ ABsin⁡3α=BCsin⁡α\frac { A B } { \sin 3 \alpha } = \frac { B C } { \sin \alpha } (Since BE = AB)

⇒ ABBC=sin⁡3αsin⁡α=3−4sin⁡2α\frac { A B } { B C } = \frac { \sin 3 \alpha } { \sin \alpha } = 3 - 4 \sin ^ { 2 } \alpha

⇒ABBC=\frac { A B } { B C } = 3−2(1−cos⁡2α)⇒ABBC=1+2cos⁡2α3 - 2 ( 1 - \cos 2 \alpha ) \Rightarrow \frac { A B } { B C } = 1 + 2 \cos 2 \alpha