Answer
Chapter 2 - Section 2.3 Problem Set: 59 (Answer)
Refer to Figure 1
The sum of the two angles
$\angle DAF$ and $\angle CAE$
Is 3.0$^{\circ}$ + 2.2$^{\circ}$ = 5.2$^{\circ}$
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 59 (Solution)
Refer to Figure 1
$EC$ = $DF$ = $6 ft.$
In $\triangle$DAB,
$\tan \angle DAB$ = $\frac{DB}{AB}$
$\tan \angle DAB$ = $\frac{(6+54)}{54}$
$\angle DAB$ = $\tan^{-1}(\frac{60}{54})$
$\angle DAB$ = 48.0$^{\circ}$ (to the nearest tenths).
In $\triangle$FAB,
$\tan \angle FAB$ = $\frac{FB}{AB}$
$\tan \angle FAB$ = $\frac{(54)}{54}$
$\angle FAB$ = $\tan^{-1}(1)$
$\angle FAB$ = 45.0$^{\circ}$ (To the nearest tenths)
$\angle DAF$ = $\angle DAB$ - $\angle FAB$
$\angle DAF$ = 48.0$^{\circ}$ - 45.0$^{\circ}$
$\angle DAF$ = 3.0$^{\circ}$ (To the nearest tenths)
Similarly,
In $\triangle$EAB,
$\tan \angle EAB$ = $\frac{EB}{AB}$
$\tan \angle EAB$ = $\frac{(24+54)}{54}$
$\angle EAB$ = $\tan^{-1}(\frac{78}{54})$
$\angle EAB$ = 55.3$^{\circ}$ (To the nearest tenths)
In $\triangle$CAB,
$\tan \angle CAB$ = $\frac{CB}{AB}$
$\tan \angle CAB$ = $\frac{(12+6+54)}{54}$
$\angle CAB$ = $\tan^{-1}(\frac{72}{54})$
$\angle CAB$ = 53.1$^{\circ}$ (To the nearest tenths)
$\angle CAE$ = $\angle EAB$ - $\angle CAB$
$\angle CAE$ = 55.3$^{\circ}$ - 53.1$^{\circ}$
$\angle CAE$ = 2.2$^{\circ}$ (To the nearest tenths)
Therefore, the sum of the two angles
$\angle DAF$ and $\angle CAE$
Is 3.0$^{\circ}$ + 2.2$^{\circ}$ = 5.2$^{\circ}$
