C7.2; E7.2 - Finding sides with trigonometry

C7.2; E7.2 - Finding sides with trigonometry

A 10 m cable makes an angle of 38∘38^\circ with a straight baseline in a right-angled triangle. How far does it reach across that baseline? C7.2 and E7.2 connect acute angles to side ratios, so one known side and an angle can determine another side.

Trigonometric ratios

Two right-angled triangles with the same acute angle have the same shape. Enlarging one multiplies every side by the same factor, so ratios of matching sides stay unchanged. Trigonometry names three of these ratios.

Take a right triangle with legs 3 and 4 and hypotenuse 5. Choose the acute angle facing the side of length 3. Its opposite-to-hypotenuse ratio is 3/53/5. Enlarging to legs 6 and 8 and hypotenuse 10 gives 6/10=3/56/10=3/5, the same ratio.

Opposite, adjacent and hypotenuse

For a chosen acute angle θ\theta, the opposite side faces θ\theta. The adjacent side is the shorter side touching θ\theta. The hypotenuse faces the right angle and is always the longest side.

The adjacent side excludes the hypotenuse, even though both touch the acute angle. Mark the angle first, because changing the chosen acute angle exchanges opposite and adjacent.

[DIAGRAM: asset_name: Sides relative to an acute angle; asset_slug: trig_relative_sides; caption: Changing the chosen acute angle swaps opposite and adjacent, while the hypotenuse stays the same.; description: Two equal 3 by 4 by 5 right triangles ABC with right angle at B. The left highlights angle A and labels BC opposite, AB adjacent and AC hypotenuse. The right highlights angle C and labels AB opposite, BC adjacent and AC hypotenuse. Labels are placed outside the triangles.]
Changing the chosen acute angle swaps opposite and adjacent, while the hypotenuse stays the same.

Right-triangle trigonometric ratios

For an acute angle θ\theta in a right-angled triangle,

sin⁡θ=oppositehypotenusecos⁡θ=adjacenthypotenusetan⁡θ=oppositeadjacent\begin{aligned} \sin\theta&=\frac{\text{opposite}}{\text{hypotenuse}}\\ \cos\theta&=\frac{\text{adjacent}}{\text{hypotenuse}}\\ \tan\theta&=\frac{\text{opposite}}{\text{adjacent}} \end{aligned}

Not on the supplied formula list for Papers 1–4: learn these ratios. “SOH CAH TOA” records their side pairs.

For the chosen angle in the 3–4–5 triangle, sin⁡θ=3/5\sin\theta=3/5, cos⁡θ=4/5\cos\theta=4/5 and tan⁡θ=3/4\tan\theta=3/4. These are numbers, not side lengths. Sine and cosine lie between 0 and 1 for acute angles, because their numerators are shorter than the hypotenuse. Tangent can exceed 1 when opposite is longer than adjacent.

Key idea

The angle chooses the ratio; the triangle's scale chooses its actual side lengths. Two same-shaped triangles can have different lengths but identical trigonometric ratios.

Choose the ratio that uses the named pair of sides.

Choosing a ratio

Use the known side and the side you need. A ratio containing a third, unknown side adds unnecessary work. Label the three sides relative to the marked angle, then choose the formula containing exactly the required pair.

Sides involvedRatio
Opposite and hypotenuseSine
Adjacent and hypotenuseCosine
Opposite and adjacentTangent

Worked example 1: the opposite side is unknown

Question. Triangle ABCABC is right-angled at BB. AC=10 cmAC=10\,\mathrm{cm}, ∠BAC=32∘\angle BAC=32^\circ, and BC=x cmBC=x\,\mathrm{cm}. Write an equation to find xx.

Plan. Label the known side and the unknown relative to angle AA.

AC=hypotenuseopposite right angle BBC=oppositefaces angle Asin⁡32∘=x10opposite over hypotenuse\begin{aligned} AC&=\text{hypotenuse} &&\text{opposite right angle }B\\ BC&=\text{opposite} &&\text{faces angle }A\\ \sin32^\circ&=\frac{x}{10} &&\text{opposite over hypotenuse} \end{aligned}

Answer. sin⁡32∘=x/10\sin32^\circ=x/10, or x=10sin⁡32∘x=10\sin32^\circ.

Check. The fraction compares a shorter side to the hypotenuse, so it should lie between 0 and 1. Sine uses precisely those two sides.

Now choose the other acute angle in the same triangle. The side BCBC becomes adjacent, while the hypotenuse remains ACAC.

Continue this lesson.

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