in a triangle ABC ,the bisector of angle B and angle C intersect each other at point O. prove that angle BOC =90°+1/2 angle A.
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in a triangle ABC we have
angle[A+B+C] = 180 DEGREE [since,sum of the angles of a tri.]
=> 1/2[A+B+C] = 90 DEGREE
=> 1/2A + angle1+angle2 = 90 degree
=> angle1 + angle2 = 90 - 1/2angle A -------------- equation 1
NOW in trianglel OBC,
angle1 + angle2 + angle BOC = 180
90 - 1/2 angleA + angle BOC = 180 FROM 1
angle BOC = 90degree + 1/2 angle A.
angle[A+B+C] = 180 DEGREE [since,sum of the angles of a tri.]
=> 1/2[A+B+C] = 90 DEGREE
=> 1/2A + angle1+angle2 = 90 degree
=> angle1 + angle2 = 90 - 1/2angle A -------------- equation 1
NOW in trianglel OBC,
angle1 + angle2 + angle BOC = 180
90 - 1/2 angleA + angle BOC = 180 FROM 1
angle BOC = 90degree + 1/2 angle A.
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