Pythagoras links three sides. Sine, cosine and tangent link a side to an angle, so one side and one angle are enough to solve a whole right-angled triangle.
Label the sides from the angle: hypotenuse (opposite the right angle), opposite (facing the angle), adjacent (next to the angle). Then \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\).
Fig. 1: sides labelled from the 35° angle at A, drawn to scale. BC = 10 sin 35° = 5.74 cm and AB = 10 cos 35° = 8.19 cm
Section 01
Label, choose, solve.
In the lab, pick an angle and watch the three sides relabel themselves as the angle moves to the other corner. Then change the angle and see the ratio opposite ÷ hypotenuse match the sine button on your calculator.
Step 1, label: mark the hypotenuse first, then opposite and adjacent from the angle you are using.
Step 2, choose: tick the two sides involved (the one you know and the one you want). SOH, CAH or TOA is the ratio that uses exactly those two.
Step 3, solve: if the unknown is on top, multiply (\(x = 10\sin 35^\circ\)). If the unknown is on the bottom, divide (\(x = \dfrac{6}{\cos 52^\circ}\)).
Calculator: must be in degree mode (a small D on the display).
Worked example Unknown on top
In Fig. 1, \(AC = 10\) cm and angle \(A = 35^\circ\). Find \(BC\).
From angle \(A\): \(AC\) is the hypotenuse and \(BC\) is opposite. Opposite and hypotenuse means SOH.
\(\sin 35^\circ = \dfrac{BC}{10}\).
Multiply both sides by 10: \(BC = 10\sin 35^\circ = 5.7357\ldots\)
Answer: \(BC = 5.74\) cm (3 s.f.)
Worked example Unknown on the bottom
A right-angled triangle has an angle of \(52^\circ\). The side adjacent to this angle is 6 cm. Find the hypotenuse \(h\).
Adjacent and hypotenuse means CAH: \(\cos 52^\circ = \dfrac{6}{h}\).
Multiply by \(h\): \(h\cos 52^\circ = 6\).
Divide by \(\cos 52^\circ\): \(h = \dfrac{6}{\cos 52^\circ} = 9.7456\ldots\)
Answer: \(h = 9.75\) cm (3 s.f.). Check: longer than 6 cm, as a hypotenuse must be.
Examiner trap
Writing \(h = 6\cos 52^\circ = 3.69\) cm. When the unknown is on the bottom of the fraction you divide, not multiply. A 'hypotenuse' shorter than the other side is a sure sign of this slip.
Section 02
Finding an angle.
When two sides are known, find the ratio, then use the inverse key (\(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\), usually SHIFT then sin, cos or tan) to turn the ratio back into an angle. Give angles to 1 decimal place.
Worked example Inverse tangent
In a right-angled triangle, the side opposite angle \(x\) is 7 cm and the side adjacent to it is 4 cm. Find \(x\).
Opposite and adjacent means TOA: \(\tan x = \dfrac{7}{4} = 1.75\).
\(x = \tan^{-1}(1.75) = 60.255\ldots^\circ\)
Answer: \(x = 60.3^\circ\) (1 d.p.)
Worked example Angles in an isosceles triangle
An isosceles triangle has equal sides of 12 cm and a base of 10 cm. Find the size of each base angle.
The line of symmetry splits it into two right-angled triangles with hypotenuse 12 cm and base \(10 \div 2 = 5\) cm.
From the base angle \(\theta\), 5 cm is adjacent and 12 cm is the hypotenuse: \(\cos\theta = \dfrac{5}{12}\).
In a right-angled triangle the hypotenuse is 5.8 cm and the side opposite angle \(y\) is 3.2 cm. Find \(y\).
Opposite and hypotenuse: \(\sin y = \dfrac{3.2}{5.8}\), so \(y = \sin^{-1}(0.5517\ldots) = 33.5^\circ\) (1 d.p.).
Examiner trap
Using the full base (10 cm) in the isosceles triangle gives \(\cos^{-1}(10/12) = 33.6^\circ\). The right angle is only in the half-triangle, so use half the base.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
You know the side opposite an angle and the hypotenuse. Which ratio links them?
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A right-angled triangle has an angle of \(40^\circ\) with an adjacent side of 8 cm. Find the opposite side \(x\).
Quick check+10 XP
In a right-angled triangle the adjacent side is 5 cm and the hypotenuse is 9 cm. Find the angle.
Quick check+10 XP
In triangle \(PQR\), angle \(Q = 90^\circ\). Using angle \(P\), which side is the adjacent?
Section 03
Problems in two dimensions.
Real problems often join two right-angled triangles along a common side. Find the shared side in the first triangle, keep its full value in your calculator, then use it in the second.
Worked example Two triangles, one shared side
In triangle \(ABC\), \(D\) is the point on \(BC\) with \(AD\) perpendicular to \(BC\). \(AB = 9\) cm, angle \(ABC = 40^\circ\) and angle \(ACB = 25^\circ\). Find \(AD\) and \(DC\).
Triangle \(ABD\): \(AB\) is the hypotenuse and \(AD\) is opposite the \(40^\circ\), so \(AD = 9\sin 40^\circ = 5.785\ldots\) cm.
Triangle \(ADC\): from the \(25^\circ\) angle, \(AD\) is opposite and \(DC\) is adjacent, so \(\tan 25^\circ = \dfrac{AD}{DC}\).
\(DC = \dfrac{5.785\ldots}{\tan 25^\circ} = 12.406\ldots\) cm
Answer: \(AD = 5.79\) cm, \(DC = 12.4\) cm (3 s.f.)
Worked example Angle of a diagonal
A rectangle is 15 cm long and 8 cm wide. Find the angle between a diagonal and a longer side.
The diagonal makes a right-angled triangle with sides 15 cm (adjacent to the angle) and 8 cm (opposite).
\(\tan\theta = \dfrac{8}{15}\), so \(\theta = \tan^{-1}(0.5333\ldots) = 28.07\ldots^\circ\)
Answer: \(28.1^\circ\) (1 d.p.)
Examiner trap
Rounding \(AD\) too early. With \(AD = 5.8\) you only just survive (\(DC = 12.44\)); with \(AD = 6\) you get \(DC = 12.9\) cm and lose the accuracy mark. Keep the unrounded value in your calculator.
Stage 2 · Exam
Exam-style questions
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A straight ramp is 4.5 m long and rises a vertical height of 0.6 m. Find the angle the ramp makes with the horizontal.
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In a right-angled triangle, the side opposite an angle of \(28^\circ\) is 9 cm. Find the hypotenuse.
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An isosceles triangle has two sides of 12 cm and a base of 10 cm. Find the angle between the two equal sides.
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\(AD\) is the perpendicular height of triangle \(ABC\), with \(D\) on \(BC\). \(AB = 9\) cm, angle \(B = 40^\circ\), angle \(C = 25^\circ\). Find \(BC\).
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SOH CAH TOA, sorted.
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