Question : A pole stands vertically on a road, which goes in the north-south direction. P and Q are two points towards the north of the pole, such that $P Q=b$, and the angles of elevation of the top of the pole at $P, Q$ are $\alpha, \beta$ respectively. Then the height of the pole is:
Option 1: $\frac{b}{\tan \beta+\tan \alpha}$
Option 2: $\frac{b}{\tan \beta-\tan \alpha}$
Option 3: $\frac{b}{\cot \beta-\cot \alpha}$
Option 4: $\frac{\mathrm{b} \tan \alpha}{\tan \beta}$
Correct Answer: $\frac{b}{\cot \beta-\cot \alpha}$
Solution :
Let the height of the pole be $h$ cm.
$PQ = b$
In $\triangle ABQ$
⇒ $\tan \beta = \frac{AB}{BQ}$
⇒ $\tan \beta = \frac{h}{BQ}$
⇒ $BQ = \frac{h}{\tan \beta}$
⇒ $BQ = h \cot \beta$
In $\triangle ABP$
⇒ $\tan \alpha = \frac{AB}{BP}$
⇒ $\tan \alpha = \frac{h}{BP}$
⇒ $BP = \frac{h}{\tan \alpha}$
⇒ $BP = h \cot \alpha$
⇒ $BQ = BP + PQ$
⇒ $h \cot \beta = h \cot \alpha + b$
⇒ $h (\cot \beta - \cot \alpha) = b$
⇒ $h = \frac{b}{\cot \beta - \cot \alpha}$
Hence, the correct answer is $\frac{b}{\cot \beta - \cot \alpha}$.
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