JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q89

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MathematicsSingle correct+4 / -1Question 89 of 90

The angle of elevation of the top of a vertical tower from a point P on the horizontal ground was observed to be $\alpha$ . After moving a distance 2 metres from P towards the foot of the tower, the angle of elevation changes to $\beta$ . Then the height (in metres) of the tower is :

\dfrac{2\sin\alpha\sin\beta}{\sin(\beta-\alpha)}

\dfrac{\sin\alpha\sin\beta}{\cos(\beta-\alpha)}

\dfrac{2\sin(\beta-\alpha)}{\sin\alpha\sin\beta}

\dfrac{\cos(\beta-\alpha)}{\sin\alpha\sin\beta}

Answer and worked solution

Official answer: A

$h\cot\alpha-h\cot\beta=2$ , and $\cot\alpha-\cot\beta=\dfrac{\sin(\beta-\alpha)}{\sin\alpha\sin\beta}$ . So
$h=\frac{2\sin\alpha\sin\beta}{\sin(\beta-\alpha)}.$

Topics:Heights and distances

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JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 89