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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Question : A clock tower stands at the crossing of two roads which point in the north-south and the east-west directions. $P, Q, R$, and $S$ are points on the roads due north, east, south, and west respectively, where the angles of elevation of the top of the tower are respectively,
Question : The angles of elevation of a pole from two points which are 75 m and 48 m away from its base are $\alpha$ and $\beta$, respectively. If $\alpha$ and $\beta$ are complementary, then the height of the tower is:
Question : If $\sin \beta=\frac{1}{3},(\sec \beta-\tan \beta)^2$ is equal to:
Question : If $\alpha$ is an acute angle and $2\sin \alpha+15\cos^2\alpha=7$, then the value of $\cot \alpha$ is:
Question : If $\sec A - \tan A = p$, then find the value of $\sec A$.
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