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Height and Distance
Height and Distance
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Question

The angle of elevation of the top of a tower from a point A due north of it is α\alpha and from a point B at a distance of 9 units due west of A is cos⁡−1(313)\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right). If the distance of the point B from the tower is 15 units, then cot⁡α\cot \alpha is equal to :

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Solution

NA=152−92=12NA = \sqrt {{{15}^2} - {9^2}} = 12 h15=tan⁡θ=23{h \over {15}} = \tan \theta = {2 \over 3} h=10h = 10 units cot⁡α=1210=65\cot \alpha = {{12} \over {10}} = {6 \over 5}

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