Chapter 11.2, Problem 36STP

To determine

To find: The value of $x$

Answer to Problem 36STP

  $x^0={56}^0$

Explanation of Solution

Given information:

Angles of a triangle are ${90}^0,{34}^0,x^0$

Angles of a triangle are ${90}^0,{34}^0,x^0$

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

Using that

We get

  ${90}^0+{34}^0+x^0={180}^0$

On solving

We get

  $\begin{array}{l}{90}^0+{34}^0+x^0={180}^0\\ \Rightarrow {124}^0+x^0={180}^0\\ \Rightarrow x^0={180}^0-{124}^0\\ \Rightarrow x^0={56}^0\end{array}$

Therefore,

  $x^0={56}^0$