In the diagram, a wooden ramp is shown where $\angle BCF = 90^\circ$ and $\angle CBF = 35^\circ$. ABCD, CDEF and ABFE are rectangles. AB = 40 cm and BC = 75 cm. Find $\angle FAC$. Answer ___

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 $ACF$,

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

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

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