Use sine, cosine and tangent to find unknown sides and angles in right-angled triangles, including angles of elevation and depression

What this dot point is asking

SEAB wants you to use the sine, cosine and tangent ratios to find unknown sides and angles in right-angled triangles, and to apply them to angles of elevation and depression in real situations. Choosing the correct ratio from the sides involved is the central skill.

The answer

The three ratios

In a right-angled triangle, label the sides relative to a chosen angle: the hypotenuse (opposite the right angle), the opposite (across from the angle), and the adjacent (next to the angle). The three ratios are:

The memory aid SOH CAH TOA captures all three.

Choosing the right ratio

Identify which two sides are involved (one known, one wanted), then pick the ratio that uses exactly those two sides. If the hypotenuse and opposite appear, use sine; adjacent and hypotenuse, cosine; opposite and adjacent, tangent.

Finding a side and finding an angle

To find a side, substitute the angle and known side, then rearrange. To find an angle, form the ratio of the two known sides and apply the inverse function ($\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$) on the calculator, in degree mode.

Angles of elevation and depression

The angle of elevation is measured upward from the horizontal to a line of sight; the angle of depression is measured downward from the horizontal. These angles, with a horizontal or vertical distance, form a right-angled triangle to which the ratios apply.

Examples in context

Example 1. Height of a building. Standing a known distance away and measuring the angle of elevation to the top lets you compute a building's height with the tangent ratio, without climbing it. Surveyors use this routinely.

Example 2. Navigation and slopes. The angle of depression from a clifftop to a boat, with the cliff height, gives the boat's distance from the base by the tangent ratio. The same idea finds gradients of roads and ramps.

Try this

Q1. In a right-angled triangle the adjacent side is $8\ \text{cm}$ and the angle is $40^\circ$. Find the hypotenuse, to 2 decimal places. [2 marks]

• Cue. $\cos 40^\circ = \dfrac{8}{\text{hyp}}$, so hyp $= \dfrac{8}{\cos 40^\circ} = 10.44\ \text{cm}$.

Q2. The opposite side is $6\ \text{cm}$ and the hypotenuse is $10\ \text{cm}$. Find the angle, to 1 decimal place. [2 marks]

• Cue. $\sin\theta = \dfrac{6}{10} = 0.6$, so $\theta = \sin^{-1}(0.6) = 36.9^\circ$.

Q3. State which ratio links the opposite side and the adjacent side. [1 mark]

• Cue. Tangent, since $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$.