In the diagram, PQRS is a vertical rectangular wall of height 2.5 m and length 7.5 m. T is a point on level ground from the base of the wall RS. ∠STR = 95° and ∠SRT = 40°. X is a point on ST such that SX : XT is 1 : 3.

1. Find the remaining angle in triangle $SRT$.

In $\triangle SRT$,
$\angle RST=180^\circ-95^\circ-40^\circ=45^\circ.$

2. Use the sine rule to calculate $ST$.

Using the sine rule,
$\frac{ST}{\sin 40^\circ}=\frac{SR}{\sin 95^\circ}.$
Since $SR=7.5$,
$ST=\frac{7.5\sin 40^\circ}{\sin 95^\circ}=4.839\ldots\text{ m}.$

3. Use the ratio $SX:XT=1:3$ to find $SX$.

Since $SX:XT=1:3$,
$SX=\frac{1}{1+3}\times ST=\frac14(4.839\ldots)=1.209\ldots\text{ m}.$

1. Identify the included angle in triangle $SXR$.

From triangle $SRT$,
$\angle RST = 180^\circ - 95^\circ - 40^\circ = 45^\circ.$
Since $X$ lies on $ST$,
$\angle XSR = 45^\circ.$

2. Apply the cosine rule to triangle $SXR$.

Using $SR=7.5$ m and $SX=1.21$ m from part (a),
$XR^2 = SX^2 + SR^2 - 2(SX)(SR)\cos \angle XSR$
$XR^2 = 1.21^2 + 7.5^2 - 2(1.21)(7.5)\cos45^\circ.$

3. Evaluate and give the length of $XR$.

$XR^2 \approx 44.88$
$XR \approx \sqrt{44.88} = 6.70.$

1. Find $ST$ from the given ratio and earlier result for $SX$.

Since $SX:XT=1:3$, $SX$ is $\dfrac14$ of $ST$.

Using $SX=1.21\text{ m}$,
$ST=4(1.21)=4.84\text{ m}.$

2. Apply Pythagoras' theorem to find $PT$.

In right-angled triangle $PST$,
$PT^2=PS^2+ST^2$
$PT=\sqrt{2.5^2+4.84^2}=5.446\ldots$

So, to 3 significant figures,
$PT=5.45\text{ m}.$

1. Identify where the bird should be for the angle of elevation to be largest.

The bird is on $PQ$, which is vertically above $SR$ by $2.5\text{ m}$.

Since the height is fixed, $\theta$ is largest when the horizontal distance from $X$ to the bird's vertical projection on $SR$ is smallest.

So the bird should be vertically above the foot of the perpendicular from $X$ to $SR$.

2. Find the shortest horizontal distance from $X$ to $SR$.

From triangle $SRT$,
$\angle RST = 180^\circ - 95^\circ - 40^\circ = 45^\circ.$

Since $X$ lies on $ST$, $\angle XSR = 45^\circ$.

Using $SX = 1.21\text{ m}$ from part (a),
$\text{horizontal distance} = 1.21\sin 45^\circ$
$= 0.8556\text{ m}.$

3. Use tangent ratio to find the maximum angle of elevation.

In the vertical right-angled triangle from $X$ to the bird,
$\tan \theta = \frac{2.5}{0.8556}.$

$\theta = \tan^{-1}\left(\frac{2.5}{0.8556}\right)$
$\theta = 71.1^\circ \text{ correct to 1 decimal place}.$