Question

In figure angle ACD is an exterior angle of triangle ABC, if angle B = 40 degree, angle A = 70 degree find angle ACD ​

Answer 1

Answer:

Given, ∠A = 70° and ∠B = 40°

In a triangle ABC,

The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles

∠ACD is an exterior angle of triangle ABC

So, from theorem of remote interior angles,

∠ACD = ∠BAC + ∠ABC

⇒ ∠ACD = ∠A + ∠B

⇒ ∠ACD = 70° + 40° = 110°

Answer 2

Answer:

$\huge\color{blue}\boxed{\colorbox{lightgreen}{♧Solution♧}}\bigstar$

We are given m∠B=40°

m∠A=70°

Now,

BD is a straight ray

so, m∠ACB+m∠ACD=180°→(1)

$In △ABC,m∠A+m∠B+m∠ACB=180 {}^{0} →(2)$

From (1) & (2)

m∠A+m∠B=m∠ACD

so, m∠ACD=40° +70°

, m∠ACD=40° +70°

m∠ACD=110°