# Chapter 3 - Section 3.1 - Reference Angle - 3.1 Problem Set - Page 122: 82

$315^{\circ}$

#### Work Step by Step

Given- $\sec\theta$ = $\sqrt 2$ , $\theta$ belongs to Q IV We will find reference angle of $\theta$ first using value $\sqrt 2$ and reference angle theorem- $\sec\theta$ = $\sqrt 2$ = $\sec 45^{\circ}$ (As $\sec 45^{\circ}$ = $\sqrt 2$) Therefore by reference angle theorem, $45^{\circ}$ is the reference angle for desired angle $\theta$ The desired angle $\theta$ is in Q IV, Also $0^{\circ}\leq \theta\lt 360^{\circ}$ Therefore- $\theta$ = $360^{\circ} - 45^{\circ}$ = $315^{\circ}$

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