How do I find a missing angle in a right-angled triangle?

Work out the ratio of the two sides you know, then use the inverse function. With the opposite 5 cm and the hypotenuse 10 cm, that is opposite ÷ hypotenuse = 5 ÷ 10 = 0.5, which is a sine, so the angle is sin⁻¹(0.5) = 30°. Using cos⁻¹(0.5) = 60° would be the wrong ratio for those two sides. Match the ratio to the sides, then invert.

When do I add and when do I subtract in Pythagoras?

Add the squares only when you are finding the hypotenuse, the longest side opposite the right angle. When you are finding a shorter side you subtract: shorter side = √(hypotenuse² − other shorter side²). With hypotenuse 13 cm and one shorter side 12 cm, the missing shorter side is √(13² − 12²) = √(169 − 144) = √25 = 5 cm. Adding gives √(13² + 12²) = √313 ≈ 17.7 cm, which is longer than the 13 cm hypotenuse — impossible, and the giveaway that you added when you should have subtracted.

GCSE Maths Foundation · AQA · Angles, shape & trigonometry

Trigonometry & Pythagoras: pick the right ratio, subtract for a shorter side

Right-angled triangles trip students two ways. First, they reach for sin, cos or tan without first labelling which side is opposite, adjacent or hypotenuse, so they pick a ratio that does not use the sides they have. Second, in Pythagoras they add the squares for every missing side, when finding a side shorter than the hypotenuse needs a subtraction.

The thirty-second fix: label opposite / adjacent / hypotenuse first, then pick the ratio that uses your two sides; and in Pythagoras add the squares only for the hypotenuse, subtract them for a shorter side. The opposite of a 10 cm hypotenuse at 35° is $10 \sin 35^\circ \approx 5.7$ cm, and a shorter side is $\sqrt{13^2 - 12^2} = \sqrt{25} = 5$ cm.

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How to spot it in your own work

You wrote down sin, cos or tan before labelling the opposite, adjacent and hypotenuse — so the ratio uses sides you do not have.
You inverted the wrong ratio for a missing angle, e.g. $\cos^{-1}(0.5) = 60^\circ$ where the two sides were opposite and hypotenuse, so it should be $\sin^{-1}(0.5) = 30^\circ$ .
You added the squares for a side shorter than the hypotenuse, writing $\sqrt{13^2 + 12^2}$ instead of $\sqrt{13^2 - 12^2}$ .
Your shorter side came out longer than the hypotenuse (e.g. 17.7 cm against a 13 cm hypotenuse) — impossible, the sign that you added when you should have subtracted.

An exam question that triggers it

Here is a canonical AQA Foundation trigger (calculator paper, find a missing side):

A right-angled triangle has a hypotenuse of 10 cm. The angle between the hypotenuse and the base is 35°.

Work out the length of the side opposite the 35° angle.

The misconception is to grab a ratio before labelling, or to use cos and find the adjacent side by mistake. Label first: you have the hypotenuse and want the opposite, and the ratio that uses those two is sin.

So $\text{opposite} = 10 \times \sin 35^\circ \approx 5.7$ cm. Choosing cos there would give the adjacent side, not the opposite.

Why students fall for this

The three ratios are learned as a chant — “SOH CAH TOA” — but the chant only works once the triangle’s sides are labelled relative to the angle. Skip the labelling and the chant has nothing to attach to, so the student picks whichever ratio they remember first. The opposite, adjacent and hypotenuse are defined by where the angle is, not by where the sides sit on the page.

Pythagoras has its own version of the slip. The headline form, $a^2 + b^2 = c^2$ , has a plus sign in it, so students add the squares every time. But that plus is only for finding $c$ , the hypotenuse. Rearranged for a shorter side it becomes a minus: $a^2 = c^2 - b^2$ . The hypotenuse is the longest side, so a shorter side must come out smaller than it — and adding the squares always makes the answer too big.

AQA Foundation papers exploit both directly: finding a missing side or angle with the ratios (label opp/adj/hyp, then pick sin/cos/tan), and finding a shorter side with Pythagoras where $\sqrt{13^2 + 12^2}$ is the trap and $\sqrt{13^2 - 12^2}$ is the answer.

Worked example — a missing angle. A right-angled triangle has the side opposite an unknown angle measuring 5 cm, and a hypotenuse of 10 cm. Find the angle.

You have the opposite and the hypotenuse, so the ratio is sin. Form the ratio, then invert:

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{10} = 0.5 \;\Rightarrow\; \theta = \sin^{-1}(0.5) = 30^\circ

The trap is to use $\cos^{-1}(0.5) = 60^\circ$ : that is the wrong ratio for an opposite and a hypotenuse. Match the ratio to the two sides you have, then invert.

The fix: Label opp / adj / hyp first, pick the matching ratio; add the squares only for the hypotenuse

Label the three sides relative to the angle. Opposite is across from the angle; adjacent is next to it (but not the hypotenuse); the hypotenuse is opposite the right angle and is the longest side.

Pick the ratio that uses your two sides. sin uses opposite and hypotenuse, cos uses adjacent and hypotenuse, tan uses opposite and adjacent. With hypotenuse 10 and the opposite wanted at 35°, use sin: $10 \times \sin 35^\circ \approx 5.7$ cm.

For a missing angle, invert the same ratio. Opposite 5 over hypotenuse 10 is a sine of 0.5, so $\theta = \sin^{-1}(0.5) = 30^\circ$ , not $\cos^{-1}(0.5) = 60^\circ$ .

In Pythagoras, add for the hypotenuse, subtract for a shorter side. A side shorter than the hypotenuse is $\sqrt{c^2 - b^2}$ , so $\sqrt{13^2 - 12^2} = \sqrt{25} = 5$ cm. If a shorter side comes out longer than the hypotenuse, you added when you should have subtracted.

Worked example

A right-angled triangle has a hypotenuse of 13 cm and one shorter side of 12 cm. Find the length of the other shorter side. This is the add-versus-subtract trap: the missing side is shorter than the hypotenuse, so you subtract.

Spot which side is missing. The 13 cm is the hypotenuse (the longest side, opposite the right angle); the missing side is one of the shorter sides.
Subtract the squares, not add. A shorter side is $\sqrt{c^2 - b^2}$ :
$\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm}$
Sanity-check against the hypotenuse. 5 cm is shorter than the 13 cm hypotenuse, as a shorter side must be. Good.
Read off the trap. Adding the squares gives $\sqrt{13^2 + 12^2} = \sqrt{313} \approx 17.7$ cm, which is longer than the 13 cm hypotenuse — impossible. That is the tell that the plus should have been a minus.

The same labelling settles the ratio question: hypotenuse 10, opposite wanted at 35°, so use sin and $10 \times \sin 35^\circ \approx 5.7$ cm — not cos, which would have found the adjacent side instead.

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Common questions

How do I choose between sin, cos and tan?: Label the three sides relative to the angle first: opposite (across from the angle), adjacent (next to it, not the hypotenuse) and hypotenuse (opposite the right angle, the longest side). Then pick the ratio that uses the two sides you have or want — sin uses opposite and hypotenuse, cos uses adjacent and hypotenuse, tan uses opposite and adjacent. With a hypotenuse of 10 cm and the opposite unknown at 35°, you have hypotenuse and want opposite, so use sin: $10 \times \sin 35^\circ \approx 5.7$ cm. Choosing cos there would give the adjacent side, not the opposite.
How do I find a missing angle in a right-angled triangle?: Work out the ratio of the two sides you know, then use the inverse function. With the opposite 5 cm and the hypotenuse 10 cm, that is $\tfrac{5}{10} = 0.5$ , which is a sine, so the angle is $\sin^{-1}(0.5) = 30^\circ$ . Using $\cos^{-1}(0.5) = 60^\circ$ would be the wrong ratio for those two sides. Match the ratio to the sides, then invert.
When do I add and when do I subtract in Pythagoras?: Add the squares only when you are finding the hypotenuse, the longest side opposite the right angle. When you are finding a shorter side you subtract: $\sqrt{c^2 - b^2}$ . With hypotenuse 13 cm and one shorter side 12 cm, the missing shorter side is $\sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ cm. Adding gives $\sqrt{13^2 + 12^2} = \sqrt{313} \approx 17.7$ cm, longer than the 13 cm hypotenuse — impossible, and the giveaway that you added when you should have subtracted.

Related misconceptions

Angles in polygons: exterior angles sum to 360°The neighbouring shape fact: each exterior angle of a regular n-gon is 360 ÷ n, and more sides means a smaller angle, not a bigger one.
Symmetry & 2D/3D shape: lines of symmetry and projectionsReading a shape by its real properties, not its appearance: counting lines of symmetry by folding and drawing plan and elevation as flat 2D views.
Triangle area: base × height ÷ 2, not base × heightAnother right-triangle slip: a triangle is half its bounding rectangle, so its area is base × perpendicular height ÷ 2.

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