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QJEEM22Ma2dc1: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2dc1 Solution: The angle of elevation of the top P of a vertical tower PQ of…

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∘45^{\circ} . Let R be a point on AQ and from a point B, vertically above R\mathrm{R} , the angle of elevation of P\mathrm{P} is 60∘60^{\circ} . If ∠BAQ=30∘,AB=d\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d} and the area of the trapezium PQRB\mathrm{PQRB} is α\alpha , then the ordered pair (d,α)(\mathrm{d}, \alpha) is :

Multiple Choice Options

  • A (10(3−1),25)(10(\sqrt{3}-1), 25)
  • B (10(3−1),252)\left(10(\sqrt{3}-1), \frac{25}{2}\right)
  • C (10(3+1),25)(10(\sqrt{3}+1), 25)
  • D (10(3+1),252)\left(10(\sqrt{3}+1), \frac{25}{2}\right)

Step-by-Step Verified Solution

Let BR=xBR = x

JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Height and Distance Question 6 English Explanation

xd=12⇒x=d2{x \over d} = {1 \over 2} \Rightarrow x = {d \over 2}

10−x10−x3=3⇒10−x=103−3x{{10 - x} \over {10 - x\sqrt 3 }} = \sqrt 3 \Rightarrow 10 - x = 10\sqrt 3 - 3x

2x=10(3−1)2x = 10(\sqrt 3 - 1)

x=5(3−1)x = 5(\sqrt 3 - 1)

d=2x=10(3−1)d = 2x = 10(\sqrt 3 - 1)

α=12(x+10)(10−x3)=\alpha = {1 \over 2}(x + 10)(10 - x\sqrt 3 )= Area (PQRB)

=12(53−5+10)(10−53(3−1))= {1 \over 2}\left( {5\sqrt 3 - 5 + 10} \right)\left( {10 - 5\sqrt 3 (\sqrt 3 - 1)} \right)

=12(53+5)(10−15+53)−12(75−25)=25= {1 \over 2}\left( {5\sqrt 3 + 5} \right)\left( {10 - 15 + 5\sqrt 3 } \right) - {1 \over 2}(75 - 25) = 25

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