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(D) Let $\overline{AB}$ be the flag-staff on top of the building $\overline{BC}$ and $D$ be the point of observation.Given,$CD = 15\, m$.In $\Delta BC

Vedclass · Accepted Answer

$10.95$

Vedclass · Answer

(D) Let $\overline{AB}$ be the flag-staff on top of the building $\overline{BC}$ and $D$ be the point of observation.Given,$CD = 15\, m$.In $\Delta BCD$,$\angle C = 90^{\circ}$ and $\angle BDC = 45^{\circ}$.$	an(45^{\circ}) = \frac{BC}{CD} \implies 1 = \frac{BC}{15} \implies BC = 15\, m$.In $\Delta ACD$,$\angle C = 90^{\circ}$ and $\angle ADC = 60^{\circ}$.$	an(60^{\circ}) = \frac{AC}{CD} \implies \sqrt{3} = \frac{AC}{15} \implies AC = 15\sqrt{3} \approx 15 	imes 1.732 = 25.98\, m$.The height of the flag-staff $AB = AC - BC = 25.98 - 15 = 10.98\, m$. Using $\sqrt{3} \approx 1.73$,$AC = 25.95\, m$,so $AB = 25.95 - 15 = 10.95\, m$.

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