Chapter 12.3, Problem 64WE

To determine

To find all angles $\theta ,0^{\circ }\le \theta <{360}^{\circ }$ , that makes the given statement true.

Answer to Problem 64WE

  $\theta ={40}^{\circ },{140}^{\circ }$

Explanation of Solution

Given information :

  $\sin \theta =\sin {40}^{\circ }$

  $\begin{array}{l}\sin \theta =\sin {40}^{\circ }\\ \theta ={\sin }^{-1}\left(\sin {40}^{\circ }\right)\end{array}$

We know that the value of sin is positive inI and II quadrant.

Hence the values here can be fromI and II quadrant.

We may write $\sin {40}^{\circ }=\sin \left({180}^{\circ }-{40}^{\circ }\right)={140}^{\circ }$

Hence the values in the given range are,

  $\theta ={40}^{\circ },\left({180}^{\circ }-{40}^{\circ }\right)={140}^{\circ }$