SSC CPO SI and ASI 2012 Paper

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Question : 135 of 200

Marks: +1, -0

Internal bisectors of ∠B and ∠C of a triangle ABC meet at O. If ∠BAC = 80°, then the value of ∠BOC is

120°
140°
110°
130°

Solution:

(4)

∠ B+\∠ C=180^{\°}-∠A

=180^{\°}-80^{\°}=100^{\°}

S o, {∠ B}/{2}+ {∠ C}/{2}=50^{\°} \

and

hence,

\∠ O B C+\∠ O C B=50^{\°}

Now, in ΔBOC,

\∠ B O C=180^{\°}-50^{°}=130^{\°}

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