KVPY SX SB Exam 2016 Question Paper

Question : 10 of 120

Marks: +1, -0

A sphere with centre O sits atop a pole as shown in the figure. An observer on the ground is at a distance 50m from the foot of the pole. She notes that the angles of elevation from the observer to points P and Q on the sphere are 30º and 60º, respectively. Then, the radius of the sphere in meters is

$100(1-1/√3)$
${50√6}/3$
$50(1-1/√3)$
${100√6}/3$

Validate

Solution:

DE = BC = r

\tan 30°=h/50

⇒h=50/√3

\tan 60°={h+r}/{50-r}

⇒√3(50-r)=h+r

⇒√3(50-r)=50/√3+r

⇒100=(3+√3)r

⇒r={100/{3+√3}}=50(1-1/√3)

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