Chapter 11.2, Problem 11IP

To determine

To find: Value of $x$ in each triangle.

Answer to Problem 11IP

  $x={70}^0$

Explanation of Solution

Given information:

Two angles of a triangle are ${55}^0$ and ${55}^0$

Corresponding sides of a triangle are $2.8cm$ and $2.8cm$

Given

Two angles of a triangle are ${55}^0$ and ${55}^0$

We know sum of angles of a triangle is ${180}^0$

Using that

We get

  $x^0+{55}^0+{55}^0={180}^0$

On solving

  $\begin{array}{l}{x}^0+{55}^0+{55}^0={180}^0\\ \Rightarrow x^0+{110}^0={180}^0\\ \Rightarrow x^0={180}^0-{110}^0\\ \Rightarrow x^0={70}^0\end{array}$

Therefore,

  $x={70}^0$

As two angles and their corresponding sides are equal and all angles are less than ${90}^0$

So, it is an Acute isoceless angled triangle