Solve the following:

In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

Dear Student

From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
= b + b + 40^o = 180^o
= 2b + 40^o = 180^o
By transposing 40 from LHS to RHS it becomes -40
= 2b = 180–40
= 2b = 140
Now,
Divide both the side by 2
= 2b/2 = 140/2
= b = 70^o
Hence, 70^ois the base angle of an isosceles triangle.

Thanks