ABL is a park on horizontal ground. A is 1390 m from L on a bearing of 076°. B is 960 m from L on a bearing of 138°.

1. Find the included angle ALB using the two bearings from L.

Bearing of A from L $=076^\circ$

Bearing of B from L $=138^\circ$

Hence
$\angle ALB=138^\circ-76^\circ=62^\circ.$

2. Apply the cosine rule to find AB.

Using the cosine rule,
$AB^2=1390^2+960^2-2(1390)(960)\cos62^\circ.$

$AB=\sqrt{1390^2+960^2-2(1390)(960)\cos62^\circ}$

$AB=1241.5\ldots$

So, to 3 significant figures,
$AB=1240\text{ m}.$

1. Identify the known angle at $L$ and set up the sine rule.

From the bearings,
$\angle ALB = 138^\circ - 76^\circ = 62^\circ.$
Using $AB=1240\text{ m}$ from part (a)(i),
$\frac{\sin \angle LAB}{LB}=\frac{\sin \angle ALB}{AB}$
$\frac{\sin \angle LAB}{960}=\frac{\sin 62^\circ}{1240}.$

2. Calculate angle $LAB$.

$\sin \angle LAB=\frac{960\sin62^\circ}{1240}$
$\angle LAB=\sin^{-1}\left(\frac{960\sin62^\circ}{1240}\right)$
$\angle LAB=43.1^\circ\text{ }\text{(to 3 s.f.)}.$

1. Find the included angle between $LA$ and $LB$.

$\angle ALB=138^\circ-76^\circ=62^\circ$

2. Apply the area formula using two sides and the included angle.

$\text{Area of }\triangle LAB=\frac12 ab\sin C$
$=\frac12(1390)(960)\sin 62^\circ$
$=589122.\ldots$

1. Find the bearing of L from A using the reverse bearing of A from L.

The bearing of $A$ from $L$ is $076^\circ$.

So the bearing of $L$ from $A$ is

$076^\circ + 180^\circ = 256^\circ.$

2. Use the angle at A to find the bearing of B from A.

From part (a)(ii),

$\angle LAB = 43.1^\circ.$

At $A$, the direction from $A$ to $B$ is clockwise less than the direction from $A$ to $L$, so

$\text{bearing of }B\text{ from }A = 256^\circ - 43.1^\circ = 212.9^\circ.$

Correct to the nearest degree,

$212.9^\circ \approx 213^\circ.$

1. Find the shortest horizontal distance from L to the path AB.

The greatest angle of depression occurs at the point on $AB$ closest to $L$.

This shortest distance is the perpendicular distance from $L$ to $AB$.

Using the area from part (a)(iii),
$\text{Area of }\triangle LAB=\frac12\times AB\times d$
where $d$ is the perpendicular distance from $L$ to $AB$.

$589000=\frac12\times 1240\times d$
$d=\frac{2\times 589000}{1240}=950\text{ m}$

2. Set up the right-angled trigonometric ratio for the tower height.

Let the height of the tower be $h$ metres.

The angle of depression is equal to the angle of elevation from the nearest point on $AB$ to $T$.

$\tan 5.06^\circ=\frac{h}{950}$

3. Calculate the height of the tower.

$h=950\tan 5.06^\circ$
$h=84.1\text{ m} \quad \text{to 3 s.f.}$

1. Find the total distance Jonah jogs around the edge of the park.

Using $LA=1390\text{ m}$, $AB=1240\text{ m}$ from part (a)(i), and $BL=960\text{ m}$,

$\text{total distance}=1390+1240+960=3590\text{ m}.$

2. Use speed to calculate the time taken.

Convert $3590\text{ m}$ to kilometres:

$3590\text{ m}=3.59\text{ km}.$

At $8.5\text{ km/h}$,

$\text{time}=\frac{3.59}{8.5}\text{ h}=0.4223529\ldots\text{ h}.$

3. Convert the time to minutes and seconds and round to the nearest 10 seconds.

$0.4223529\ldots\times 60=25.34117\ldots\text{ min}.$

This is

$25\text{ min }+0.34117\ldots\times 60\text{ s}=25\text{ min }20.470\ldots\text{ s}.$

Corrected to the nearest $10$ seconds:

$25\text{ min }20\text{ s}.$