A tower subtends angles `alpha, 2alpha and 3alpha` respectively at points, A, B and C (all points lying on the same side on a horizontal line through the foot of the tower), then the value of `(AB)/(BC)` is equal to

Options

A. `1+2cos 2alpha`
B. `1-2cos2alpha`
C. `1+3cos 2alpha`
D. `1-3cos2alpha`

Answer

To solve the problem, we need to find the ratio \( \frac{AB}{BC} \) where a tower subtends angles \( \alpha, 2\alpha, \) and \( 3\alpha \) at points \( A, B, \) and \( C \) respectively. We will denote the height of the tower as \( h \) and the foot of the tower as point \( Q \).

### Step 1: Define the distances from the foot of the tower
- Let \( QA \) be the distance from point \( Q \) to point \( A \).
- Let \( QB \) be the distance from point \( Q \) to point \( B \).
- Let \( QC \) be the distance from point \( Q \) to point \( C \).

### Step 2: Express \( QA, QB, \) and \( QC \) using trigonometric functions
Using the cotangent function for the angles subtended at points \( A, B, \) and \( C \):
- From triangle \( PQA \):
\[ QA = h \cot \alpha \]
- From triangle \( PQB \):
\[ QB = h \cot 2\alpha \]
- From triangle \( PQC \):
\[ QC = h \cot 3\alpha \]

### Step 3: Calculate \( AB \) and \( BC \)
Now, we can express \( AB \) and \( BC \):
- The distance \( AB \) is given by:
\[ AB = QA - QB = h \cot \alpha - h \cot 2\alpha = h (\cot \alpha - \cot 2\alpha) \]
- The distance \( BC \) is given by:
\[ BC = QB - QC = h \cot 2\alpha - h \cot 3\alpha = h (\cot 2\alpha - \cot 3\alpha) \]

### Step 4: Set up the ratio \( \frac{AB}{BC} \)
Now we can set up the ratio:
\[ \frac{AB}{BC} = \frac{h (\cot \alpha - \cot 2\alpha)}{h (\cot 2\alpha - \cot 3\alpha)} = \frac{\cot \alpha - \cot 2\alpha}{\cot 2\alpha - \cot 3\alpha} \]

### Step 5: Simplify using trigonometric identities
To simplify further, we can use the identity for cotangent:
\[ \cot x = \frac{\cos x}{\sin x} \]
Thus, we can express the cotangent differences in terms of sine and cosine:
\[ \frac{\cot \alpha - \cot 2\alpha}{\cot 2\alpha - \cot 3\alpha} = \frac{\frac{\cos \alpha}{\sin \alpha} - \frac{\cos 2\alpha}{\sin 2\alpha}}{\frac{\cos 2\alpha}{\sin 2\alpha} - \frac{\cos 3\alpha}{\sin 3\alpha}} \]

### Step 6: Use the sine subtraction formula
Using the sine subtraction formula:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
We can express the numerator and denominator:
- For the numerator:
\[ \sin 2\alpha \cos \alpha - \cos 2\alpha \sin \alpha \]
- For the denominator:
\[ \sin 3\alpha \cos 2\alpha - \cos 3\alpha \sin 2\alpha \]

### Step 7: Final simplification
After simplification, we find:
\[ \frac{AB}{BC} = \frac{\sin(2\alpha - \alpha)}{\sin(3\alpha - 2\alpha)} = \frac{\sin \alpha}{\sin \alpha} = 1 \]

Thus, the final value of \( \frac{AB}{BC} \) is:
\[ \frac{AB}{BC} = 1 \]