The diagram shows a rectangular table ABCD arranged such that the side AD is inclined at an angle of $\theta$ to the wall. The lengths of AB and BC are 0.7 m and 1.8 m respectively. The perpendicular distance from C to the wall is $L$ metres.

1. Find the perpendicular component contributed by side $BC$.

Since $AD$ is inclined at angle $\theta$ to the wall and $BC\parallel AD$, side $BC$ also makes angle $\theta$ with the wall.

Hence the component of $BC$ perpendicular to the wall is
$1.8\sin\theta.$

2. Find the perpendicular component contributed by side $AB$ and add the components.

Since $AB\perp BC$, the component of $AB$ perpendicular to the wall is
$0.7\cos\theta.$

Therefore the total perpendicular distance from $C$ to the wall is
$L=1.8\sin\theta+0.7\cos\theta.$

1. Expand the required form using the cosine compound-angle formula.

$R\cos(\theta-\alpha)=R(\cos\theta\cos\alpha+\sin\theta\sin\alpha)$
$=R\cos\alpha\cos\theta+R\sin\alpha\sin\theta$

2. Compare coefficients with $1.8\sin\theta+0.7\cos\theta$ and find $R$.

Comparing
$R\cos\alpha\cos\theta+R\sin\alpha\sin\theta\equiv 0.7\cos\theta+1.8\sin\theta,$
we get
$R\cos\alpha=0.7,\qquad R\sin\alpha=1.8.$
Hence
$R^2=(R\cos\alpha)^2+(R\sin\alpha)^2=0.7^2+1.8^2=3.73,$
so
$R=\sqrt{3.73}.$

3. Find the acute angle $\alpha$.

$\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha}=\frac{1.8}{0.7}$
$\alpha=\tan^{-1}\left(\frac{1.8}{0.7}\right)=68.7^\circ \text{, to 3 s.f.}$
Therefore,
$L=\sqrt{3.73}\cos(\theta-68.7^\circ).$

1. Use the maximum value of the cosine function to identify the maximum possible value of $L$.

From part (ii),
$L=\sqrt{3.73}\cos(\theta-68.7^\circ).$
Since the maximum value of $\cos x$ is $1$,
$L_{\max}=\sqrt{3.73}\times 1=\sqrt{3.73}.$

2. Find the value of $\theta$ that gives the maximum value of $L$.

For the maximum value,
$\cos(\theta-68.7^\circ)=1.$
Thus
$\theta-68.7^\circ=0^\circ,$
so
$\theta=68.7^\circ.$

1. Substitute $L=\sqrt{2.5}$ into the form found in part (ii).

From part (ii),
$L=\sqrt{3.73}\cos(\theta-68.7^\circ).$
When $L=\sqrt{2.5}$,
$\sqrt{2.5}=\sqrt{3.73}\cos(\theta-68.7^\circ)$
$\cos(\theta-68.7^\circ)=\frac{\sqrt{2.5}}{\sqrt{3.73}}.$

2. Use inverse cosine to find the possible angle, then choose the value of $\theta$ consistent with the diagram.

$\theta-68.7^\circ=\pm \cos^{-1}\left(\frac{\sqrt{2.5}}{\sqrt{3.73}}\right)$
$\cos^{-1}\left(\frac{\sqrt{2.5}}{\sqrt{3.73}}\right)\approx 35.0^\circ$
So
$\theta=68.7^\circ-35.0^\circ=33.7^\circ$
or
$\theta=68.7^\circ+35.0^\circ=103.7^\circ.$
Since $\theta$ is an acute angle of inclination in the diagram,
$\theta=33.7^\circ.$