Chapter 11.2, Problem 10IP

To determine

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

Answer to Problem 10IP

  $x={37}^0$

Explanation of Solution

Given information:

Two angles of a triangle are ${120}^0$ and ${23}^0$

Given

Two angles of a triangle are ${120}^0$ and ${23}^0$

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

Using that

We get

  $x^0+{120}^0+{23}^0={180}^0$

On solving

  $\begin{array}{l}{x}^0+{120}^0+{23}^0={180}^0\\ \Rightarrow x^0+{143}^0={180}^0\\ \Rightarrow x^0={180}^0-{143}^0\\ \Rightarrow x^0={37}^0\end{array}$

Therefore,

  $x={37}^0$

As one of the angle is greater than ${90}^0$

So, it is an Obtuse angled triangle