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Cambridge IGCSE Mathematics — 0580 Extended

Topic 6.2: Trigonometry — Right-angled Triangles

Label opposite, adjacent, hypotenuse

Sine, cosine and tangent compare two sides of a right-angled triangle. The labels Opposite, Adjacent and Hypotenuse are always taken from the acute angle you are using — not from the right angle.

\[ \sin\theta = \frac{\text{O}}{\text{H}} \qquad \cos\theta = \frac{\text{A}}{\text{H}} \qquad \tan\theta = \frac{\text{O}}{\text{A}} \]

Right-angled triangle ABC with angle theta at A, opposite BC, adjacent AC and hypotenuse AB labelled, and the three SOHCAHTOA ratios
Opposite is the side that does not touch \(\theta\). Adjacent touches \(\theta\) but is not the hypotenuse. The hypotenuse is opposite the right angle.

Method

  1. Circle the angle you are using and mark the right angle.
  2. Label O, A and H from that angle. Relabel if the question switches angle.
  3. Choose the ratio that uses the two sides you care about, then rearrange.
  4. Calculator in degree mode. Decimal answers to \(1\) d.p. unless the paper says otherwise.

Finding a side

Write the ratio that contains the known side and the unknown. Rearrange so the unknown is the subject, then evaluate.

In a right-angled triangle the hypotenuse is \(12\,\text{cm}\) and one acute angle is \(35^\circ\). Find the side opposite \(35^\circ\).

Right triangle with hypotenuse 12 cm and angle 35 degrees, opposite side x found using sine as 6.9 cm
Need O and have H, so sine: \(x = 12\sin 35^\circ = 6.9\,\text{cm}\) (\(1\) d.p.).

Finding an angle

When two sides are known, form the matching ratio and take the inverse: \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\).

The side opposite \(\theta\) is \(7\,\text{cm}\) and the side adjacent to \(\theta\) is \(11\,\text{cm}\). Find \(\theta\).

Right triangle with opposite 7 cm and adjacent 11 cm, theta found using inverse tan as 32.5 degrees
Have O and A, so tangent: \(\theta = \tan^{-1}\left(\frac{7}{11}\right) = 32.5^\circ\) (\(1\) d.p.).

Angles of elevation and depression

An angle of elevation is measured up from the horizontal at the observer. An angle of depression is measured down from the horizontal at the observer. Depression at one end of a sight-line equals elevation at the other (alternate angles with the two horizontals).

From a point \(20\,\text{m}\) from the base of a tree, the angle of elevation of the top is \(32^\circ\). Find the height of the tree.

Observer O, 20 metres from tree base B, elevation 32 degrees to the top T, height 12.5 metres
\(\tan 32^\circ = \frac{h}{20}\), so \(h = 20\tan 32^\circ = 12.5\,\text{m}\) (\(1\) d.p.).

From the top of an \(80\,\text{m}\) cliff the angle of depression of a boat is \(25^\circ\). Find the distance of the boat from the base of the cliff.

Cliff 80 metres high, dashed horizontal through the observer, depression 25 degrees to a boat, distance 171.6 metres from the base
The dashed line is the horizontal through the observer. \(\tan 25^\circ = \frac{80}{x}\), so \(x = \frac{80}{\tan 25^\circ} = 171.6\,\text{m}\) (\(1\) d.p.).

Bearings

A bearing is an angle measured clockwise from north and written with three figures (\(036.9^\circ\), not \(36.9^\circ\)). Drop perpendiculars to make a right-angled triangle, then use \(\tan\) (or Pythagoras) as usual.

Point \(C\) is \(8\,\text{km}\) due north and \(6\,\text{km}\) due east of \(A\). Find the bearing of \(C\) from \(A\).

North arrow at A, point C 8 km north and 6 km east, right triangle 6-8-10, bearing of C from A equal to 036.9 degrees
\(\tan\theta = \frac{6}{8}\), so \(\theta = 36.9^\circ\). Clockwise from north, the bearing is \(036.9^\circ\). Distance \(AC = 10\,\text{km}\) by Pythagoras.

Shortest distance to a line

The perpendicular from a point to a line is the shortest distance. Any other path from the point to the line is the hypotenuse of a right-angled triangle, so it is longer. To find that distance, drop a perpendicular and then use Pythagoras or trigonometry.

Point P above line AB with perpendicular foot F marked shortest, and two dashed longer slanted paths from P to the line
\(PF \perp AB\) is shorter than every other segment from \(P\) to \(AB\).

Try this

A ladder of length \(10\,\text{m}\) leans against a vertical wall. The angle between the ladder and the ground is \(55^\circ\). How far is the foot of the ladder from the wall?

Show answer
Answer

  1. The ground is adjacent to \(55^\circ\); the ladder is the hypotenuse

    \[ \cos 55^\circ = \frac{x}{10} \]

  2. Degree mode, \(1\) d.p.

    \[ x = 10\cos 55^\circ = 5.7\,\text{m} \]

Exam Traps

  • Elevation and depression are measured from the horizontal, never from the vertical wall or cliff face. A dashed horizontal through the observer is worth drawing.
  • If \(\sin 32^\circ\) comes out near \(0.53\) you are in degree mode; a result near \(0.55\) (or a huge number) usually means the calculator is in radians.

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