图中显示了一个木制斜坡，其中 $\angle BCF = 90^\circ$ 且 $\angle CBF = 35^\circ$。ABCD、CDEF 和 ABFE 均为矩形。AB = 40 cm 且 BC = 75 cm。求 $\angle FAC$。 答案 ___

1. Find the diagonal $AC$ of rectangle $ABCD$.

Since $ABCD$ is a rectangle, $\angle ABC=90^\circ$.

$AC=\sqrt{AB^2+BC^2}=\sqrt{40^2+75^2}=\sqrt{7225}=85\text{ cm}$

2. Find the perpendicular height $CF$ using triangle $BCF$.

In right-angled triangle $BCF$,

$\tan 35^\circ=\frac{CF}{BC}$

$CF=75\tan 35^\circ$

3. Use triangle $ACF$ to find $\angle FAC$.

In right-angled triangle $ACF$,

$\tan \angle FAC=\frac{CF}{AC}$

$\tan \angle FAC=\frac{75\tan 35^\circ}{85}$

$\angle FAC=\tan^{-1}\left(\frac{75\tan 35^\circ}{85}\right)=31.7^\circ\text{, to 3 s.f.}$