Chapter 11.2, Problem 6IP

To determine

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

Answer to Problem 6IP

  $x={77}^0$

Explanation of Solution

Given information:

Two angles of a triangle are ${50}^0$ and ${53}^0$

Given

Two angles of a triangle are ${50}^0$ and ${53}^0$

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

Using that

We get

  $x^0+{50}^0+{53}^0={180}^0$

On solving

  $\begin{array}{l}{x}^0+{50}^0+{53}^0={180}^0\\ \Rightarrow x^0+{103}^0={180}^0\\ \Rightarrow x^0={180}^0-{103}^0\\ \Rightarrow x^0={77}^0\end{array}$

Therefore,

  $x={77}^0$

As all angles are less than ${90}^0$

So, it is an Acute angled triangle