KVPY 2019 SB SX Exam Previous Papers

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Question : 82 of 120

Marks: +1, -0

Let ABC be a triangle and let D be the midpoint of BC. Suppose cot(∠CAD) : cot(∠BAD) = 2 : 1. If G is the centroid of triangle ABC, then the measure of ∠BGA is

90º
105º
120º
135º

Solution:

Given

{ \cotα}/{\cotβ}=2/1

{ \cotα}-2{\cotβ}

⇒

{AG^2+ AE^2- GE^2}/{2(AG)(AE)\sinα}=2[{AB^2+AG^2-BG^2}/{2(AB)(AG)\sinβ}]

⇒

{{2b^2+2c^2-a^2}/9+b^2/4-{2a^2+2c^2-b^2}/36}/{4/3\ar(∆ADC) }

={c^2+{2b^2+2c^2-a^2}/9 -{2a^2+2c^2-b^2}/9}/{4/3\ar(∆ABD) }

⇒

4(2b^2 + 2c^2 -a^2) + 9b^2 -2a^2 -2c^2 + b^2

= 36c^2 + 4(3b^2 -3a^2)

⇒

a^2 + b^2 = 5c^2

\cosθ ={AG^2+BG^2-c^2}/{2(AG)(BG)}

{AG^2+BG^2-c^2}

= {{2b^2+2c^2-a^2}/9 +{2a^2+2c^2-b^2}/9} -c^2

{a^2+ b^2- 5c^2}/9=0⇒θ = 90º

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