A tower subtends angles `alpha,2alpha,3alpha` respectively, at point `A , B ,a n dC` all lying on a horizontal line through the foot of the tower. Prove that `(A B)/(B C)=1+2cos2alphadot`

Options

A. `(3 sin alpha)/(sin 2 alpha)`
B. `1+2 cos ^2 alpha`
C. `2+cos^3 alpha`
D. `(sin 2 alpha)/(sin alpha )`

Explanation

To solve the problem, we need to prove that \(\frac{AB}{BC} = 1 + 2 \cos 2\alpha\). Let's break down the solution step by step.

### Step 1: Understanding the Angles
Let \(P\) be the foot of the tower. The angles subtended by the tower at points \(A\), \(B\), and \(C\) are \(\alpha\), \(2\alpha\), and \(3\alpha\) respectively.

### Step 2: Applying the Exterior Angle Theorem
In triangle \(PAB\), by the exterior angle theorem, we have:
\[
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