## Trigonometry 7th Edition

Chapter 2 - Section 2.3 Problem Set: 27 (Answer) Refer to Figure I $\angle B$ = $79^{\circ}18{}’$ $c = 6.037$ cm (To four significant digits) $a = 1.121$ cm (To four significant digits)
Chapter 2 - Section 2.3 Problem Set: 27 (Solution) Refer to Figure I $\angle A$ + $\angle B$ + $90^{\circ}$ = $180^{\circ}$ ($\angle$ sum of $\triangle$) $10.7^{\circ}$ + $\angle B$ + $90^{\circ}$ = $180^{\circ}$ ($10^{\circ}42{}'$ = $10\frac{42}{60}^{\circ}$ = $10.7^{\circ}$) $\angle B$ = $79.3^{\circ}$ = $79^{\circ}18{}’$ $\cos 10.7^{\circ} = \frac{b}{c}$ $c = \frac{5.932}{\cos 10.7^{\circ}}$ $c = 6.037$ cm (To four significant digits) $\tan 10.7^{\circ} = \frac{a}{b}$ $a = 5.932 \cdot \tan 10.7^{\circ}$ $a = 1.121$ cm (To four significant digits)