Question

In an isosceles triangle$ABC$, the angle $A$at the vertex is$70°$. Find the base angles $\angle Band\angle C$ taking $AB=AC$

Answer 1

Given, in a $△ABC$

$$\begin{gathered} AB=AC \\ and\angle A=70° \end{gathered}$$

We have to find $\angle Band\angle C$

We know that, by the property of the triangle

If $AB=AC$ then $\angle B=\angle C$

Now, we also know that from the property of the triangle

$\angle A+\angle B+\angle C=180°$

So,

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow 70°+\angle B+\angle B=180°(\sin ce\angle A=70°and\angle B=\angle C) \\ \Rightarrow 70°+2\angle B=180° \\ \Rightarrow 2\angle B=180°-70°=110 \\ \Rightarrow \angle B=\frac{110°}{2}=55° \end{gathered}$$

Since, $\angle B=\angle C$

So,$\angle C=55°$

Hence, the base angles $\angle B=55°and\angle C=55°$.