Question 805747: 1. in a triangle ABC angle=80 degree, if BD and CD are internal bisectors of angle B and angle C respectively, then the angle BDC is,
2. the side BC of the triangle ABC is produced to D. if angle ACD=112 degree and angle ABC=3/4 angle BAC, then the angle ABC is equal to
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! 1. in a triangle ABC angle A=80 degree, if BD and CD are internal bisectors of angle B and angle C respectively, then the angle BDC is,
In triangle ABC angle A + Angle B +angle C = 180 Deg
Angle B +angle C = 180 Deg- angle A
=180-80=100 deg
In triangle BDC angle DBC + angle DCB + BDC = 180 deg
angle DBC = 1/2 angle B ( angular bisectors)
angle DCB = 1/2 angle C
In triangle BDC 1/2 angle B +1/2 angle C + BDC = 180 deg
In triangle BDC 1/2 (angle B + angle C) + BDC = 180 deg
=1/2(100)+BDC= 180 deg
angle BDC = 180 -50
angle BDC= 130 deg
2. the side BC of the triangle ABC is produced to D. if angle ACD=112 degree and angle ABC=3/4 angle BAC, then the angle ABC is equal to
Let angle BAC = x
angle ABC = 3x/4
x+ (3x/4) =112 ( The exterior angle is equal to the sum of two interior remote angles)
7x/4=112
7x=112*4
x=16*4
x=64 deg (BAC)
ABC = 3*64/4
ABC = 48 deg
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