KVPY 2020-21 SA Exam Previous Papers

Question : 12 of 80

Marks: +1, -0

In a triangle ABC, the altitude AD and the median AE divide ∠A into three equal parts. If BC = 28, then the nearest integer to AB + AC is

Solution:

Let AD = h
and BD = x

\ΔA B D: \tan \θ={x}/{h}

…(1)
and in

\ΔA D E: \tan \θ={14-x}/{h}

From equation (1) and (2):

{x}/{h}={14-x}/{h} \⇒x=7

Now in

\ΔA D C: \tan 2 \θ={28-x}/{h}

and in

\ΔA B D: \tan \θ={x}/{h}

\∴{\tan 2 \θ}/{\tan \θ}={28-x}/{x}

\;\;\;⇒3-3 \tan ^{2} \θ=2

\∴\tan ^{2} \θ={1}/{3}

\∴\cos ^{2} \θ={3}/{4}

and

\sin ^{2} \θ={1}/{4}

Here

A B={7}/{\sin \θ}

and

A C={14}/{\tan \θ}

\∴A B+A C={7}/{{1}/{2}}+{14}/{{1}/{\√{3}}}

=14(1+\√{3})=38.248

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