Answer
Chapter 2 - Section 2.3 Problem Set: 47 (Answer)
Refer to Figure I
$\angle ABD$ = $42^{\circ}$ (To the nearest digit)
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 47 (Solution)
Refer to Figure I
In $\triangle$ABC,
$\sin 41^{\circ} = \frac{x}{32}$
$x = 32 \cdot \sin 41^{\circ}$
$x = 21$ (To the nearest digit)
In $\triangle$DAB,
$\tan(\angle ABD)$ = $\frac{h}{x}$
$\tan(\angle ABD)$ = $\frac{19}{21}$
$\angle ABD$ = $\tan^{-1}(\frac{19}{21})$
$\angle ABD$ = $42^{\circ}$ (To the nearest digit)
