Question : From two points, lying on the same horizontal line, the angles of elevation of the top of the pillar are $\theta$ and $\phi$ ($\theta<\phi$). If the height of the pillar is $h$ m and the two points lie on the same sides of the pillar, then the distances between the two points are:
Option 1: $h(\tan\theta-\tan\phi)$ metre
Option 2: $h(\cot\phi-\cot\theta)$ metre
Option 3: $h(\cot\theta-\cot\phi)$ metre
Option 4: $h\frac{(\tan\theta \tan\phi)}{(\tan\phi-\tan\theta)}$ metre
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Correct Answer: $h(\cot\theta-\cot\phi)$ metre
Solution : Let AB = height of pole = $h$ metre $\angle$ACB = $\theta$, $\angle$ADB = $\phi$ In ∆ABD, $\tan\phi=\frac{AB}{BD}$ ⇒ $BD=h\cot\phi$ In ∆ABC, $\tan\theta=\frac{AB}{BC}$ ⇒ $BC=h\cot\theta$ ∴ Required distance, CD $=h\cot\theta-h\cot\phi$ $= h(\cot\theta-\cot\phi)$ metre Hence, the correct answer is $h(\cot\theta-\cot\phi)$ metre.
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Question : The angle of elevation of the top of a tower, vertically erected in the middle of a paddy field, from two points on a horizontal line through the foot of the tower are given to be $\alpha$ and $\beta(\alpha>\beta)$. The height of the tower is $h$ units. A possible distance
Question : The expression $(\tan \theta+\cot \theta)(\sec \theta+\tan \theta)(1-\sin \theta), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $(\sin \theta+\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=$?
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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