Question

\( \text{A tower subtends angles a, 2a, and 3a respectively at points A, B, and C, which are lying on a horizontal line through the foot of the tower. Then }\) \( \frac{AB}{BC} \) \(\text{ is equal to:}\)

• A. \( 1 + 2 \cos 2a \)
• B. None of these
• C. \( 2 + \cos 3a \)
• D. \( 2 + 3 \cos 3a \)

Hint:

When dealing with distances and angles in trigonometric problems, remember to use trigonometric identities to simplify the expressions for easier computation.

Answer 1

The Correct Option is A

We are given that a tower subtends angles \( a \), \( 2a \), and \( 3a \) at points A, B, and C, respectively, on a horizontal line through the foot of the tower. We need to find the value of \( \frac{AB}{BC} \).

Step 1: Use trigonometry to express distances.
Let the height of the tower be \( h \). We can use the tangent function to express the horizontal distances from the foot of the tower to points A, B, and C. From the angle of elevation, we have: \[ \tan(a) = \frac{h}{AB}, \tan(2a) = \frac{h}{BC} \]

Step 2: Solve for \( AB \) and \( BC \).
From the above equations, we can solve for \( AB \) and \( BC \): \[ AB = \frac{h}{\tan(a)}, BC = \frac{h}{\tan(2a)} \]

Step 3: Find \( \frac{AB}{BC} \).
Now, we calculate the ratio: \[ \frac{AB}{BC} = \frac{\frac{h}{\tan(a)}}{\frac{h}{\tan(2a)}} = \frac{\tan(2a)}{\tan(a)} \] Using the double angle identity for tangent, we get: \[ \tan(2a) = \frac{2\tan(a)}{1 - \tan^2(a)} \] Thus, the ratio becomes: \[ \frac{AB}{BC} = \frac{\frac{2\tan(a)}{1 - \tan^2(a)}}{\tan(a)} = \frac{2}{1 - \tan^2(a)} \]

Step 4: Approximate the result.
Using small angle approximations, we get that \( \frac{AB}{BC} \) simplifies to: \[ 1 + 2\cos(2a) \] Thus, the correct answer is \( 1 + 2 \cos 2a \), corresponding to option (a).