Question

In the given figure,∆ACB is a right angled triangle at C and CD AB . If angle B =40° , then find.​

Answer 1

Step-by-step explanation:

in triangle bcd angle b +c +d=180°

90+40+c=180

c=50°

90°=angle bcd +acd

90-50=acd

acd=40°

in triangle ABC

90°+40°+angle a=180

angle a=50°

Answer 2

Step-by-step explanation:

<BCD+BDC+DBC=180°

<BCD+90°+40°=180°

<BCD= 180°-130°

<BCD= 50°

<ACD=<BCD

<ACD=50°

<BAC+<ACB+<ABC=180

<BAC +100°+40°= 180°

<BAC = 180°-140°

<BAC= 40°

1. <BAC=40°
2. <ACD=50°
3. <BCD=50°