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Height and Distance
Height and Distance
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Question

A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = 5\sqrt 5 inches and ∠\anglePCB = tan -1 (2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is :

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Solution

From figure, sin⁡β=15\sin \beta = {1 \over {\sqrt 5 }} ∴\therefore tan⁡β=12\tan \beta = {1 \over 2} tan⁡(α+β)=2\tan (\alpha + \beta ) = 2 tan⁡α+tan⁡β)1−tan⁡α.tan⁡β=2{{\tan \alpha + \tan \beta )} \over {1 - \tan \alpha .\tan \beta }} = 2 tan⁡α+121−tan⁡α(12)=2{{\tan \alpha + {1 \over 2}} \over {1 - \tan \alpha \left( {{1 \over 2}} \right)}} = 2 tan⁡α=34\tan \alpha = {3 \over 4} α=tan⁡−1(34)\alpha = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)

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