Question

in triangleABC , sides BC is produced to D . if angle ABC = 40 and angle ACD= 120 find angle A

Answer 1

Answer:

The correct answer is $\angle A =80^{\circ}$.

Explanation:

Given:

In triangle ABC , sides BC exists produced to D . if angle ABC = 40 and angle ACD = 120.

To find:

angle A

Step 1

Let, $\angle A B C=40^{\circ}$ and $\angle A CD=120^{\circ}$

then, $\angle A +\angle B=\angle ACD$

We know that the exterior angle of a triangle exists equivalent to the sum of its interior opposite angles.

$\angle \mathrm{ACD}+\angle \mathrm{ACB}=180^{\circ}$

Step 2

$\angle \mathrm{A}+40^{\circ}=120^{\circ}$

As we know that

$\angle A =80^{\circ}$

Therefore, the correct answer is $\angle A =80^{\circ}$.

#SPJ3

Answer 2

Answer:

this is the answer of this question very easy