Answer
Chapter 2 - Section 2.3 Problem Set: 60 (Answer)
Refer to Figure 1
The sum of the two angles
$\angle DAF$ and $\angle CAE$
is 6.2$^{\circ}$ + 6.2$^{\circ}$ = 12.4$^{\circ}$
Work Step by Step
Chapter 2 - Section 2.3 Problem Set: 60 (Solution)
Refer to Figure 1
EC = DF = 6 ft, M is the mid-point of CD.
In $\triangle$FAM,
$\tan \angle FAM$ = $\frac{FM}{AM}$
$\tan \angle FAM$ = $\frac{(6+6)}{54}$
$\angle FAM$ = $\tan^{-1}(\frac{12}{54})$
$\angle FAM$ = 12.5$^{\circ}$ (to the nearest tenths).
In $\triangle$DAM,
$\tan \angle DAM$ = $\frac{DM}{AM}$
$\tan \angle DAM$ = $\frac{6}{54}$
$\angle DAM$ = $\tan^{-1}(\frac{6}{54})$
$\angle DAM$ = 6.3$^{\circ}$ (to the nearest tenths).
$\angle DAF$ = $\angle FAM$ - $\angle DAM$
$\angle DAF$ = 12.5$^{\circ}$ - 6.3$^{\circ}$
$\angle DAF$ = 6.2$^{\circ}$ (to the nearest tenths).
Since $\triangle$DAF and $\triangle$CAE are similar,
$\angle DAF$ = $\angle CAE$ = 6.2$^{\circ}$ (to the nearest tenths).
Therefore, the sum of the two angles
$\angle DAF$ and $\angle CAE$
is 6.2$^{\circ}$ + 6.2$^{\circ}$ = 12.4$^{\circ}$.
