Answer
Chapter 2 - Section 2.3 Problem Set: 54 (Answer)
Refer to Figure 1
$x = 36$ (To 2 significant digits)
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 54 (Solution)
Refer to Figure 1
In $\triangle$BAC,
$\tan A$ = $\frac{h}{(x+y)}$
$\tan 32^{\circ}$ = $\frac{h}{(x+14)}$
$h = (x+14) \cdot \tan 32^{\circ}$ $\longrightarrow (1)$
In $\triangle$BDC,
$\tan \angle BDC$ = $\frac{h}{x}$
$\tan 41^{\circ}$ = $\frac{h}{x}$
$h = x \cdot \tan 41^{\circ}$ $\longrightarrow (2)$
Equating $(1) to (2)$
$(x+14) \cdot \tan 32^{\circ}$ = $x \cdot \tan 41^{\circ}$
$x = \frac{14\tan 32^{\circ}}{\tan 41^{\circ} - \tan 32^{\circ}}$
$x = 36$ (To 2 significant digits)
