Chapter 11.2, Problem 30HP

To determine

To find: Values of $x$ and $y$ in the figure

Answer to Problem 30HP

  $\begin{array}{l}x={50}^0\\ y={80}^0\end{array}$

Explanation of Solution

From the given figure

Taking larger triangle

Sum of measures of the angles of a triangle is ${180}^0$

Using that

We get

  ${35}^0+{95}^0+x^0={180}^0$

On solving

We get

  $\begin{array}{l}{35}^0+{95}^0+x^0={180}^0\\ \Rightarrow {130}^0+x^0={180}^0\\ \Rightarrow x^0={180}^0-{130}^0\\ \Rightarrow x^0={50}^0\end{array}$

Taking smaller triangle

Sum of measures of the angles of a triangle is ${180}^0$

Using that

We get

  ${50}^0+y^0+x^0={180}^0$

On solving

We get

  $\begin{array}{l}{50}^0+y^0+x^0={180}^0\\ \Rightarrow {50}^0+{50}^0+y^0={180}^0\\ \Rightarrow {100}^0+y^0={180}^0\\ \Rightarrow y^0={180}^0-{100}^0\\ \Rightarrow y^0={80}^0\end{array}$

Therefore,

  $\begin{array}{l}x={50}^0\\ y={80}^0\end{array}$