The diagram shows the positions of a harbour, H, a lighthouse, L, and two buoys P and Q. HPQ is a straight line. The bearing of P from H is 306°. HP = 3 km, PL = 2.5 km and angle HPL = 124°.

1. Find the reverse bearing from P to H

The bearing of P from H is $306^\circ$.

The bearing of H from P is the reverse bearing, so subtract $180^\circ$:
$306^\circ - 180^\circ = 126^\circ.$

1. Find the bearing of L from P using the bearing of H from P and the angle HPL

From part (a)(i), the bearing of H from P is $126^\circ$.

Given $\angle HPL = 124^\circ$, and $PL$ is on the other side of North from $PH$,
$\text{bearing of L from P}=126^\circ-124^\circ=2^\circ.$

Written as a three-figure bearing: $002^\circ$.

1. Use the tangent ratio to find the angle of elevation

Let the angle of elevation be $\theta$. Since $PL=2.5\text{ km}=2500\text{ m}$,
$\tan \theta=\frac{500}{2500}=0.2$
$\theta=\tan^{-1}(0.2)=11.3^\circ \text{ correct to 3 s.f.}$

1. Apply the cosine rule to triangle HPL

In $\triangle HPL$, $HP=3$, $PL=2.5$ and $\angle HPL=124^\circ$.

$HL^2=HP^2+PL^2-2(HP)(PL)\cos \angle HPL$

$HL^2=3^2+2.5^2-2(3)(2.5)\cos124^\circ$

2. Calculate and give the length HL

$HL^2=9+6.25-15\cos124^\circ$

$HL=\sqrt{9+6.25-15\cos124^\circ}$

$HL=4.852\ldots$

So, $HL\approx 4.85\text{ km}$

1. Find the distance from P to the nearest point X on the line HPQ

Since $X$ is the point on $HPQ$ nearest to $L$, $LX \perp HPQ$.

As $PQ$ is the extension of $PH$,
$\angle LPX=180^\circ-124^\circ=56^\circ.$

In right-angled triangle $PLX$,
$\cos 56^\circ=\frac{PX}{PL}$
$PX=2.5\cos56^\circ=1.39798\ldots\text{ km}.$

2. Find the total distance sailed from H to X

$HX=HP+PX$
$HX=3+1.39798\ldots=4.39798\ldots\text{ km}$
$=4397.98\ldots\text{ m}.$

3. Use speed to find the travel time and subtract it from 0809 hours

Average speed $=4.5\text{ m/s}$.

$\text{time}=\frac{4397.98\ldots}{4.5}=977.33\ldots\text{ s}$
$977.33\ldots\text{ s}=16.288\ldots\text{ min}$

The ship reached $X$ at $0809$ hours, so it left about $16$ minutes earlier.

$0809-16\text{ min}=0753.$