Chapter 1.6, Problem 1PSA

To determine

To show: $\angle V\cong \angle S$.

Answer to Problem 1PSA

  $\angle V=\angle S$ .

Explanation of Solution

Given information:

A figure is given as below. $\angle V=119{\frac{2}{3}}^0,\angle S={119}^0{40}^{\text{'}}$ .

  

Calculation:

Given, $\angle V=119{\frac{2}{3}}^0,\angle S={119}^0{40}^{\text{'}}$ .

Since one degree has ${60}^{\text{'}}\text{ i}\text{.e}\text{. }{1}^0={60}^{\text{'}}$ .

  $\text{So, }{\frac{2}{3}}^0=\frac{2}{3}\times {60}^{\text{'}}={40}^{\text{'}}$ .

  $\begin{array}{l}\text{Thus, }\angle V=119{\frac{2}{3}}^0={119}^0{40}^{\text{'}},\angle S={119}^0{40}^{\text{'}}\\ \Rightarrow \angle V=\angle S\end{array}$