Angles of Elevation, Depression, and Bearings (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Helicopter Landing Pad

A helicopter at point $B$ observes a landing pad at point $D$. The horizontal distance is $9$ units and the direct distance is $19$ units. Find the angle of depression from the helicopter to the landing pad.

Solution approach:

First, identify which angle we need to find. The angle of depression is labelled $\angle CBD$ in the diagram.

We know:

• Adjacent side $BC = 9$
• Hypotenuse $BD = 19$
• Unknown angle $x$

Since we have the adjacent side and hypotenuse, we use the cosine ratio:

$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

Substituting our values:

$\cos x = \frac{9}{19}$

To find $x$, we apply the inverse cosine function:

$x = \cos^{-1}\left(\frac{9}{19}\right)$

Using a calculator:

$x = 61.73°$

Therefore, the angle of depression is 61.73° (to two decimal places).

Worked Example: Rocky Cliff and Boat

From the top of a rocky ledge $188$ m high, the angle of depression to a boat is $13°$. If the boat is $d$ m from the foot of the cliff, determine $d$ rounded to two decimal places.

Solution approach:

Drawing a diagram helps us visualise the situation:

Notice that the angle opposite the $188$ m side is also $13°$ (alternate angles). Since we have the opposite and adjacent sides, we use the tangent ratio:

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Where $\theta = 13°$ and $\frac{O}{A} = \frac{188}{d}$

Write the formula:

$\tan 13° = \frac{188}{d}$

Multiply both sides by $d$:

$d \tan 13° = 188$

Divide both sides by $\tan 13°$:

$d = \frac{188}{\tan 13°}$

Evaluate using a calculator:

$d = 814.32$

The boat is 814.32 m from the foot of the cliff.

Worked Example: Fighter Jet Approaching Target

A fighter jet flying at an altitude of $4000$ m is approaching a target. At point $A$, the pilot measures the angle of depression to the target to be $13°$. After a minute, at point $B$, the pilot measures the angle of depression to be $16°$.

Part a: Determine the distance $AC$, rounded to the nearest metre.

Solution approach:

With respect to the given angle $13°$, the opposite side is $4000$ m, and the adjacent side is $AC$. We use the tangent ratio:

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Substitute values:

$\tan 13° = \frac{4000}{AC}$

Multiply both sides by $AC$:

$AC \times \tan 13° = 4000$

Divide both sides by $\tan 13°$:

$AC = \frac{4000}{\tan 13°}$

$AC = 17326 \text{ m}$

Part b: Determine the distance $BC$, rounded to the nearest metre.

Solution approach:

With respect to the angle $16°$, use the tangent ratio where the opposite side is $4000$ m and the adjacent side is $BC$:

$\tan 16° = \frac{4000}{BC}$

Multiply both sides by $BC$:

$BC \times \tan 16° = 4000$

Divide both sides by $\tan 16°$:

$BC = \frac{4000}{\tan 16°}$

$BC = 13950 \text{ m}$

Part c: Determine the distance covered by the jet in one minute.

Solution approach:

The distance covered is the length of $AB$, which we find by subtracting $BC$ from $AC$:

$AB = AC - BC$

$AB = 17326 - 13950$

$AB = 3376 \text{ m}$

The jet covered 3376 m in one minute.

Worked Example: Determining Compass Bearing

A compass shows point $A$ relative to point $O$. Point $A$ is $35°$ clockwise from north. Determine the compass bearing of $A$ from $O$.

Solution:

Point $A$ is positioned $35°$ clockwise from the north direction. Since the rotation is toward the east (clockwise from north moves toward east), the bearing is written as:

N35°E

This tells us to start at north, rotate $35°$ toward the east.

Worked Example: Compass to True Bearing (Eastward)

Determine the true bearing of point $A$ from point $O$ when the compass bearing is N35°E.

Solution:

The angle between the north direction and line $OA$ measured clockwise is $35°$. Therefore, the true bearing is:

035°T

Remember to write it with three digits.

Worked Example: Compass to True Bearing (Westward)

Determine the true bearing when point $A$ is $42°$ west of north.

Solution:

To measure the angle clockwise from north when the point is to the west, we subtract the given angle from $360°$:

$360° - 42° = 318°$

The true bearing is 318°T.

Worked Example: Tower Height

A tower is $18$ m from point $P$ on the ground, making an angle of elevation of $30°$ to its top. Find the tower's exact height.

Solution approach:

We use the tangent ratio with $\theta = 30°$, where the opposite side is the height $h$ and the adjacent side is $18$ m.

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Substitute values:

$\tan 30° = \frac{h}{18}$

Multiply both sides by $18$:

$h = \tan 30° \times 18$

Use the exact value $\tan 30° = \frac{\sqrt{3}}{3}$:

$h = \frac{\sqrt{3}}{3} \times 18$

Simplify:

$h = 6\sqrt{3}$

The tower's height is $6\sqrt{3}$ m.

Worked Example: Ship's Northward Distance

A ship $S$ is $15$ km from point $P$ on a true bearing of $060°$T. Find the ship's northward distance from $P$, rounded to one decimal place.

Solution approach:

We use the cosine ratio with $\theta = 60°$. The adjacent side represents the northward distance $y$, and the hypotenuse is $15$ km.

$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

Substitute:

$\cos 60° = \frac{y}{15}$

Multiply both sides by $15$:

$y = \cos 60° \times 15$

Use the exact value $\cos 60° = \frac{1}{2}$:

$y = \frac{1}{2} \times 15$

$y = \frac{15}{2}$

$y = 7.5 \text{ km}$

The northward distance is 7.5 km.

Worked Example: Return Journey Bearing

If a hiker walks from point $A$ to point $B$ on a bearing of $120°$, determine the bearing to return to $A$ from $B$.

Solution approach:

To find the return bearing, add $180°$ to the original bearing:

$\text{Return bearing} = 120° + 180°$

$\text{Return bearing} = 300°$

The return journey bearing is 300°.

Exam tip: When the original bearing is greater than $180°$, subtract $180°$ instead of adding it. The key principle is that opposite directions are $180°$ apart.