SSC CHSL Model Paper 20

Question : 64 of 100

Marks: +1, -0

The internal bisector of angle B and angle C of a triangle ABC meet at 0. If angle A = 80°, then angle BOC is

Solution:

In triangle ABC,
A + B + C = 180
B + C = 180 - A

{\B+\C}/2=90-\A/2

In triangle OBC,
Angle BOC + angle OBC + angle OCB = 180
Angle BOC

+\B/2+\C/2=180

Angle BOC

=180-{\B+\C}/2

Angle BOC

=180-{180-\A}/2

Angle BOC

=(\angle \A/2)+90°

= 40° + 90° = 130°

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