Answer
Chapter 2 - Section 2.3 Problem Set: 31 (Answer)
Refer to Figure I
$\angle A$ = $63^{\circ}30{}’=63.5^\circ$
$c = 726$ mm (To three significant digits)
$a = 650$ mm (To three significant digits)
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 31 (Solution)
Refer to Figure I
$\angle A$ + $\angle B$ + $90^{\circ}$ = $180^{\circ}$ ($\angle$ sum of $\triangle$)
$\angle A$ + $26.5^{\circ}$ + $90^{\circ}$ = $180^{\circ}$ ($26^{\circ}30{}'$ = $26\frac{30}{60}^{\circ}$ = $26.5^{\circ}$)
$\angle A$ = $63.5^{\circ}$ = $63^{\circ}30{}’$
$\sin 26.5^{\circ} = \frac{b}{c}$
$c = \frac{324}{\sin 26.5^{\circ}}$
$c = 726$ mm (To three significant digits)
$\tan 26.5^{\circ} = \frac{b}{a}$
$a = \frac{324}{\tan 26.5^{\circ}}$
$a = 650$ mm (To three significant digits)
