SSC CHSL Model Paper 7

Question : 36 of 100

Marks: +1, -0

In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then the length of AD is

Solution:

Using the theorem of angle of bisector,

\BD/\DC=\AB/\AC=4/3

⇒

\BD =4/7x\ &\ \DC=3/7x

In ΔABD, by sine rule,

{\sin30}/{4\/7x}={\sinB}/y

...(i)
In ΔABC, by sine rule,

{\sin60}/x={\sinB}/3

√3/{2x}={\sin30.y}/{4\/7x×3}

[putting the value of sin B from (i)]
⇒

y=√3/{2x}×4/7x×3×2/1={12√3}/7

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