The height of a building is 11 , m and the he | vedclass

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(B) Let the distance between the bases of the building and the pole be $2d$. The observer is at the midpoint,so the distance from the observer to the

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$\alpha < \beta$

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(B) Let the distance between the bases of the building and the pole be $2d$. The observer is at the midpoint,so the distance from the observer to the base of the pole is $d$ and to the base of the building is $d$.For the pole of height $8 \, m$,$	an \alpha = \frac{8}{d}$.For the building of height $11 \, m$,$	an \beta = \frac{11}{d}$.Since $11 > 8$,it follows that $\frac{11}{d} > \frac{8}{d}$,which means $	an \beta > 	an \alpha$.Since the angles $\alpha$ and $\beta$ are in the first quadrant,the tangent function is strictly increasing.Therefore,$\beta > \alpha$,which is equivalent to $\alpha < \beta$.

The height of a building is $11 \, m$ and the height of a pole is $8 \, m$. Watching from the midpoint of the line segment joining their bases,the angle of elevation of the top of the pole is $\alpha$ and that of the top of the building is $\beta$. Then,$\ldots \ldots \ldots$ holds true.

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