Answer
Chapter 2 - Section 2.3 Problem Set: 48 (Answer)
Refer to Figure I
$\angle ABD$ = 66$^{\circ}$ (To the nearest digit)
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 48 (Solution)
Refer to Figure I
In $\triangle$ABC,
$\sin 49^{\circ} = \frac{x}{19}$
$x = 19 \cdot \sin 49^{\circ}$
$x = 14$ (To the nearest digit)
In $\triangle$DAB,
$\tan(\angle ABD)$ = $\frac{h}{x}$
$\tan(\angle ABD)$ = $\frac{32}{14}$
$\angle ABD$ = $\tan^{-1}(\frac{32}{14})$
$\angle ABD$ = 66$^{\circ}$ (To the nearest digit)
