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Height and Distance
Height and Distance
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Question

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 :

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Solution

Let BR=xBR = x 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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