Question 805747
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