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(D) Let the building be $BC = 10 \ m$ and the flagstaff be $AB$. Let $P$ be the point on the ground.In $\Delta BCP$,$	an 30^{\circ} = \frac{BC}{CP} =

Vedclass · Accepted Answer

$7.32$

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(D) Let the building be $BC = 10 \ m$ and the flagstaff be $AB$. Let $P$ be the point on the ground.In $\Delta BCP$,$	an 30^{\circ} = \frac{BC}{CP} = \frac{10}{CP}$.Since $	an 30^{\circ} = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{10}{CP}$,so $CP = 10\sqrt{3} \ m$.In $\Delta ACP$,$	an 45^{\circ} = \frac{AC}{CP} = \frac{AB + BC}{CP}$.Since $	an 45^{\circ} = 1$,we have $1 = \frac{AB + 10}{10\sqrt{3}}$.Therefore,$AC = 10\sqrt{3} \ m$.Length of flagstaff $AB = AC - BC = 10\sqrt{3} - 10$.$AB = 10(1.732) - 10 = 17.32 - 10 = 7.32 \ m$.

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