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Trigonometric ratios · Lesson 1 of 3

Sine, cosine and tangent

Name the sides from the angle, pick the right ratio, and use it to find a missing side or angle in a right-angled triangle.

15 minYou should already know: Angles, triangles & polygons
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Trigonometry links the angles of a right-angled triangle to the lengths of its sides. Three ratios do all the work: sine, cosine and tangent.

Name the sides from the angle

In a right-angled triangle, pick one of the two acute angles and call it θ\theta. Then:

  • the hypotenuse is the longest side, opposite the right angle;
  • the opposite side is across from θ\theta;
  • the adjacent side is next to θ\theta (and isn’t the hypotenuse).

The names depend on which angle you’re using. Change the marked angle on the board and watch the opposite and adjacent sides swap.

Naming the sidesChange the angle, then the size
35°hypotenuse 10opposite 5.7adjacent 8.2
0.574sin = opposite ÷ hypotenuse0.819cos = adjacent ÷ hypotenuse0.7tan = opposite ÷ adjacent
The hypotenuse is always the longest side, opposite the right angle. The opposite side faces the marked angle; the adjacent side runs next to it. Make the triangle bigger or smaller: the sides change but the three ratios don't. They depend only on the angle (35°), which is why a calculator can give sin 35° = 0.574.
θhypotenuseoppadjacent
Name the sides from θsin = opp ÷ hyp, cos = adj ÷ hyp, tan = opp ÷ adj

Resize the triangle on the board: the sides change, but the three ratios stay the same for the same angle. That’s why sin⁡35∘\sin 35^\circ has one value, which your calculator or tables can give you.

Finding a side

  1. Mark the angle you know and name the three sides from it.
  2. Pick the ratio that uses the side you know and the side you want.
  3. Write the equation and solve it.

Worked example · WAEC 2024

WAEC 2024 · Paper 1 · Q22

A ladder 15 m15\text{ m} long leans against a vertical pole, making an angle of 72∘72^\circ with the horizontal. Calculate, correct to one decimal place, the distance between the foot of the ladder and the pole.

  1. Draw it

    The ladder, the pole and the ground make a right-angled triangle. The ladder (15 m) is the hypotenuse, and the 72∘72^\circ angle is at the foot of the ladder, on the ground.

    Think first. The ladder is the hypotenuse. Which side is the distance from the foot of the ladder to the pole, measured from the 72° angle?

  2. Choose the ratio

    The distance to the pole runs along the ground next to the 72∘72^\circ angle: it is the adjacent side. Adjacent and hypotenuse means cosine:

    cos⁡72∘=d15\cos 72^\circ = \frac{d}{15}

  3. Solve

    d=15cos⁡72∘=15×0.3090=4.635≈4.6 md = 15\cos 72^\circ = 15 \times 0.3090 = 4.635 \approx 4.6\text{ m}

    The answer is D.

Finding an angle

If you know two sides, work out the ratio, then use the inverse function (sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1} or tan⁡−1\tan^{-1}, or the tables backwards) to get the angle.

More: sides and angles in right-angled triangles

Given one ratio, find the others

WAEC often gives one ratio as a fraction and asks for an expression in the others, without tables. Draw a right-angled triangle, put in the two sides you know, and find the third with Pythagoras.

Worked example · WAEC 2021

WAEC 2021 · Paper 1 · Q38

Given that sin⁡x=35\sin x = \frac35, 0∘≤x≤90∘0^\circ \le x \le 90^\circ, evaluate (tan⁡x+2cos⁡x)(\tan x + 2\cos x).

  1. Draw the triangle

    sin⁡x=oppositehypotenuse=35\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac35, so draw a triangle with opposite 3 and hypotenuse 5.

    Think first. sin⁡x=35\sin x = \frac35 gives the opposite and the hypotenuse. What is the adjacent side?

  2. Pythagoras for the third side

    adjacent=52−32=16=4\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4

  3. Read off the other ratios

    cos⁡x=45\cos x = \frac45 and tan⁡x=34\tan x = \frac34.

  4. Substitute

    tan⁡x+2cos⁡x=34+85=15+3220=4720=2720\tan x + 2\cos x = \frac34 + \frac85 = \frac{15 + 32}{20} = \frac{47}{20} = 2\tfrac{7}{20}

    The answer is C.

More: given one ratio, find another

The reciprocal ratios and the identities

Three more ratios are the other three turned upside down:

sec⁡θ=1cos⁡θcosec⁡θ=1sin⁡θcot⁡θ=1tan⁡θ\begin{aligned} \sec\theta &= \frac{1}{\cos\theta} \\[4pt] \operatorname{cosec}\theta &= \frac{1}{\sin\theta} \\[4pt] \cot\theta &= \frac{1}{\tan\theta} \end{aligned}

In a triangle with hypotenuse 1, the other sides are cos⁡θ\cos\theta and sin⁡θ\sin\theta, so Pythagoras gives the most useful identity of all. Dividing it by cos⁡2θ\cos^2\theta or by sin⁡2θ\sin^2\theta gives two more:

sin⁡2θ+cos⁡2θ=11+tan⁡2θ=sec⁡2θ1+cot⁡2θ=cosec⁡2θ\begin{aligned} \sin^2\theta + \cos^2\theta &= 1 \\ 1 + \tan^2\theta &= \sec^2\theta \\ 1 + \cot^2\theta &= \operatorname{cosec}^2\theta \end{aligned}

Dividing sin⁡θ\sin\theta by cos⁡θ\cos\theta gives one more: tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}, because opphyp÷adjhyp=oppadj\frac{\text{opp}}{\text{hyp}} \div \frac{\text{adj}}{\text{hyp}} = \frac{\text{opp}}{\text{adj}}.

θ1sin θcos θ
The identityLegs cos θ and sin θ, hypotenuse 1: sin²θ + cos²θ = 1

More: reciprocal ratios and identities

More: right-angled triangles inside bigger problems

The tangent also gives the angle between two lines: see the angle between two lines in coordinate geometry.

More: the angle between two lines

Your turn

WAEC 2019 · Paper 1 · Q23

In △XYZ\triangle XYZ, ∣YZ∣=32 cm|YZ| = 32\text{ cm}, ∠YXZ=52∘\angle YXZ = 52^\circ and ∠XZY=90∘\angle XZY = 90^\circ. Find, correct to the nearest centimetre, ∣XZ∣|XZ|.

32 cm52°XZY
Worked solution (try it first)
  1. YZ=32YZ = 32 cm is opposite the 52∘52^\circ angle at XX, and XZXZ is adjacent to it, so use tangent: tan⁡52∘=32XZ\tan52^\circ = \dfrac{32}{XZ}.
  2. Rearrange: XZ=32tan⁡52∘XZ = \dfrac{32}{\tan52^\circ}
    =321.280= \dfrac{32}{1.280}
    ≈25.0\approx 25.0.
  3. To the nearest centimetre, XZ=25XZ = 25 cm, option B.

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