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Mathematics Question on distance and displacement

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 45°, Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is 60°. If
∠BAQ=30°∠BAQ = 30°
, AB = d and the area of the trapezium PQRB is α, then the ordered pair (d, α) is :

A

(10(3−1),25)(10(\sqrt3-1),25)

B

(10(3−1),252)(10(\sqrt3-1),\frac{25}{2})

C

(10(3+1),25)(10(\sqrt3+1),25)

D

(10(3+1),252)(10(\sqrt3+1),\frac{25}{2})

Answer

(10(3−1),25)(10(\sqrt3-1),25)

Explanation

Solution

The correct answer is (A) : (10(3−1),25)(10(\sqrt3-1),25)
Let BR = x

Fig.

xd=12\frac{x}{d} = \frac{1}{2}
⇒x=d2⇒ x = \frac{d}{2}
10−x10−x3=3\frac{10-x}{10-x\sqrt3} = \sqrt3
⇒10−x=103−3x⇒ 10-x = 10\sqrt3-3x
2x=10(3−1)2x = 10(\sqrt3-1)
x=5(3−1)x = 5 (\sqrt3-1)
d=2x=10(3−1)d = 2x = 10(\sqrt3-1)
a=12(x+10)(10−x3)=Area( PQRB)a = \frac{1}{2} (x+10)(10-x\sqrt3) = Area (\ PQRB)
=12(53−5+10)(10−53(3−1))=\frac{1}{2} (5\sqrt3-5+10)(10-5\sqrt3(\sqrt3-1))
=12(53+5)(10−15+53)= \frac{1}{2} (5\sqrt3+5)(10-15+5\sqrt3)
=12(75−25)= \frac{1}{2}(75-25)
= 25