KVPY SA Exam 2014 Question Paper

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Question : 64 of 80

Marks: +1, -0

In a triangle ABC with

∠A = 90°,

P is a point on BC such that

PA : PB = 3 : 4

. If

AB = √7

and

AC =√5

, then BP : PC is

2 : 1
4 :3
4 : 5
8 :7

Solution:

\tan θ = √5/√7; \sin θ = √5/√12 ;

\cos θ = √7/√12

In ΔAPB using sine rule

{3x}/{\sin θ}={4x}/{\sin α}=√7/{\sin(180º -(θ+α ))}

\sin α=4/3\sinθ×√5/√12={2√5}/{3√3}

=√20/√27

⇒

\cos α = √7/√12

&

3x \sin(θ + α)= √7 \sin θ

⇒

3x (\cos α + \cot θ \sin α) =√7

⇒

3x(√7/√27+√7/√5√20/√27)=√7

⇒3x×1/√3=1

x=1/√3

∴

{BP}/{PC}={4x}/{√12-4x}=1/{{2√3}/{4\/√3}-1}

=4/{6-4}=2/1

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