In an isosceles angle ABC, the bisectors of angle B and angle C meet at a point O if angle A = 40, then angle BOC = ?
Given:
In an isosceles angle ABC, the bisectors of angle B and angle C meet at a point O if angle A = 40
To do: To find the angle BOC
Solution:
In an isosceles triangle 2 angles are equal.
Let $B = C$
$A+B+C=180°$
$40° + B + B=180°$
$40°+2B=180°$
$2B=180°-40°=140°$
$B=\frac{140°}{2}=70°$
So, $A=40°,B=C=70°$
$CO$ bisects C then $OCB=35°$. [half of C]
$OBC=35°$[half of B]
$OCB+OBC+BOC=180°$
$35°+35°+BOC=180°$
$70°+BOC=180°$
$BOC=180°-70°=110°$
Therefore, the angle of BOC is 110°
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