A vertical pole of height h stands on a build | vedclass

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Question

(A) Let the point on the ground be $O$,the foot of the building be $A$,the top of the building be $B$,and the top of the pole be $C$. Given $OA = 5$,$

Vedclass · Accepted Answer

$H^3 + hH^2 + 25H - 50h = 0$

Vedclass · Answer

(A) Let the point on the ground be $O$,the foot of the building be $A$,the top of the building be $B$,and the top of the pole be $C$. Given $OA = 5$,$AB = H$,and $BC = h$.In $	riangle OAB$,$	an \alpha = \frac{AB}{OA} = \frac{H}{5}$.Given $2 	an \beta \cot \alpha = 1$,we have $2 	an \beta = 	an \alpha = \frac{H}{5}$,so $	an \beta = \frac{H}{10}$.In $	riangle OAC$,the angle of elevation of the top of the pole is $\alpha + \beta$. Thus,$	an(\alpha + \beta) = \frac{AC}{OA} = \frac{H+h}{5}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we get:$\frac{\frac{H}{5} + \frac{H}{10}}{1 - (\frac{H}{5})(\frac{H}{10})} = \frac{H+h}{5}$$\frac{\frac{3H}{10}}{1 - \frac{H^2}{50}} = \frac{H+h}{5}$$\frac{3H}{10} \cdot \frac{50}{50 - H^2} = \frac{H+h}{5}$$\frac{15H}{50 - H^2} = \frac{H+h}{5}$$75H = (H+h)(50 - H^2)$$75H = 50H - H^3 + 50h - hH^2$$H^3 + hH^2 + 25H - 50h = 0$.

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