Chapter 8.1, Problem 1CE

To determine

To find $m\angle \text{}RSU,m\angle \text{}TSU\&m\angle \text{}T$ in the triangle shown below.

  

Answer to Problem 1CE

  $m\angle \text{}RSU={60}^{\circ },m\angle \text{}TSU={30}^{\circ }\&m\angle \text{}T={60}^{\circ }$

Explanation of Solution

Given:

  $m\angle \text{}R={30}^{\circ }$

Calculation:

In triangle RSU,

  $\begin{array}{l}m\angle \text{}R+m\angle \text{}RSU+m\angle \text{}RUS={180}^{\circ }\\ \Rightarrow {30}^{\circ }+m\angle \text{}RSU+{90}^{\circ }={180}^{\circ }\\ \Rightarrow m\angle \text{}RSU+{120}^{\circ }={180}^{\circ }\\ \Rightarrow m\angle \text{}RSU={60}^{\circ }\end{array}$

Now,

  $\begin{array}{l}m\angle \text{}RSU+m\angle \text{}TSU={90}^{\circ }\\ \Rightarrow {60}^{\circ }+m\angle \text{}TSU={90}^{\circ }\\ \Rightarrow m\angle \text{}TSU={30}^{\circ }\end{array}$

In triangle STU,

  $\begin{array}{l}m\angle \text{}SUT+m\angle \text{}TSU+m\angle \text{}T={180}^{\circ }\\ \Rightarrow {90}^{\circ }+{30}^{\circ }+m\angle \text{}T={180}^{\circ }\\ \Rightarrow {120}^{\circ }+m\angle \text{}T={180}^{\circ }\\ \Rightarrow m\angle \text{}T={60}^{\circ }\end{array}$