Answer
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)
Work Step by Step
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)
