KVPY 2016 SB Question 5

Geometry Level 4

A sphere with centre O O sits atop of a pole as shown in the figure. An observer on the ground is at a distance 50 m from the foot of the pole. The angles of elevation from the observer to the points P P and Q Q are 30 ∘ ^\circ and 60 ∘ ^\circ respectively. Find the radius of the sphere in meter.

Try my set KVPY 2016 SA Questions
100 ( 1 − 1 3 ) 100 \left(1-\frac 1{\sqrt 3} \right) 50 6 3 \frac {50\sqrt 6}3 50 ( 1 − 1 3 ) 50 \left(1-\frac 1{\sqrt 3} \right) 100 6 3 \frac {100\sqrt 6}3

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Drop a perpendicular from Q Q on A B AB , We get

Using trigonometry in Triangle A B P ABP , tan ⁡ ( 3 0 ∘ ) = A P 50 , ⋅ ⋅ ⋅ A P = 50 / 3 \tan(30^{\circ})=\frac{AP}{50} ,{\cdot}^{\cdot}\cdot AP=50/\sqrt{3}

Using trigonometry In Triangle Q C B QCB ,` tan ⁡ ( 6 0 ∘ ) = A P + R 50 − R ⋅ ⋅ ⋅ R = 100 3 ( 3 + 1 ) \tan(60^{\circ})=\frac{AP+R}{50-R} {\cdot}^{\cdot}\cdot R=\frac{100}{\sqrt{3}(\sqrt{3}+1)}

And rationalizing the denominator, we get R = 50 ( 3 − 1 ) 3 R=\frac{50(\sqrt{3}-1)}{\sqrt{3}} Which means that answer is C

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Using the Fig. by Mr. Ajinkya Shivashankar, and R as the sphere radius,
P A A B = P A 50 = T a n 30 = 1 3 , ⟹ P A = 50 3 . . . . . . ( 1 ) . \dfrac{PA}{AB}=\dfrac{PA}{50}=Tan30=\dfrac 1 {\sqrt3},~~~~\implies~PA=\dfrac{50}{\sqrt3}......(1). \\ Q C C B = O A A B − A R = P A + R 50 − R = 50 3 + R 50 − R = 3 . ⟹ 50 3 + 50 50 − R = 50 3 ∗ ( 3 + 1 ) 50 − R = 3 + 1. ∴ R = 50 ∗ ( 1 − 1 3 ) . \dfrac{QC}{CB}=\dfrac{OA}{AB-AR}=\dfrac{PA+R}{50-R}=\dfrac{\frac{50}{\sqrt3}+R}{50-R}=\sqrt3.\\ \implies~\dfrac{\frac{50}{\sqrt3}+50}{50-R}=\dfrac{\frac {50}{\sqrt3}*(\sqrt3+1)}{50-R}=\sqrt3 +1.\\ \therefore~R=50*\left (1-\frac 1 {\sqrt3} \right ).

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