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(D) Let $B$ be the base of the building,$C$ be the first floor,and $A$ be the fifteenth floor. Let $P$ be the point of observation such that $BP = 50

Vedclass · Accepted Answer

$57.5$

Vedclass · Answer

(D) Let $B$ be the base of the building,$C$ be the first floor,and $A$ be the fifteenth floor. Let $P$ be the point of observation such that $BP = 50 \ m$.In $\Delta CBP$,$\angle B = 90^{\circ}$ and $\angle CPB = 30^{\circ}$.$	an 30^{\circ} = \frac{CB}{BP}$$\frac{1}{\sqrt{3}} = \frac{CB}{50}$$CB = \frac{50}{\sqrt{3}} = \frac{50 	imes 1.732}{3} \approx 28.87 \ m$.In $\Delta ABP$,$\angle B = 90^{\circ}$ and $\angle APB = 60^{\circ}$.$	an 60^{\circ} = \frac{AB}{BP}$$\sqrt{3} = \frac{AB}{50}$$AB = 50 	imes \sqrt{3} = 50 	imes 1.732 = 86.6 \ m$.The distance between the first floor and the fifteenth floor is $AC = AB - CB$.$AC = 86.6 - 28.87 = 57.73 \ m$.Rounding to the nearest provided option,the correct answer is $57.5 \ m$.

From a point $50 \ m$ away from the base of a multistoreyed building,the angle of elevation of the first floor is found to be $30^{\circ}$ and the angle of elevation of the fifteenth floor is found to be $60^{\circ}$. What is the distance between the first floor and the fifteenth floor? (in $m$)

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