Solving for Missing Sides with Trig

When a right triangle gives you one acute angle and one side, trig finds any other side. The three ratios — sine, cosine, and tangent — each connect the angle to a specific pair of sides: $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$ . SOH-CAH-TOA keeps the pairs straight.

Every problem is the same three moves: label the sides relative to the given angle, pick the one ratio that uses your known side and your unknown side, then solve the equation. The only decision that takes thought is the labeling — after that it is arithmetic.

Label the sides first

All labels are relative to the acute angle you are working from. The hypotenuse is always the side across from the right angle — it never changes. The opposite side is the one across from your angle, not touching it. The adjacent side is the leg that touches your angle.

Switch angles and the labels switch too: the side that is opposite one acute angle is adjacent to the other. That is why labeling before choosing a ratio matters — the same side plays different roles depending on which angle the problem uses.

The figure names all three sides relative to the marked angle $\theta$ .

Pick the ratio that uses your two sides

Circle the side you know and the side you want. Opposite and hypotenuse means sine. Adjacent and hypotenuse means cosine. Opposite and adjacent means tangent. The third side of the triangle is irrelevant — ignore it.

Write the equation with the ratio's definition, for example $\sin 35^\circ = \dfrac{x}{20}$ when $x$ is opposite the $35^\circ$ angle and $20$ is the hypotenuse.

Multiply or divide?

If the unknown is on top of the fraction, multiply: from $\sin 35^\circ = \dfrac{x}{20}$ you get $x = 20\sin 35^\circ$ . If the unknown is on the bottom, you divide: from $\sin 28^\circ = \dfrac{9}{x}$ you get $x = \dfrac{9}{\sin 28^\circ}$ .

This is the step where most points are lost. A quick size check catches it: the hypotenuse is the longest side, so if you solved for a hypotenuse and got something smaller than the given leg, you multiplied when you should have divided.

Worked examples

Example 1: unknown on top (sine)

In a right triangle, the hypotenuse is $20$ and one acute angle measures $35^\circ$ . Find $x$ , the side opposite the $35^\circ$ angle, to the nearest tenth.

Opposite and hypotenuse means sine

\sin 35^\circ = \dfrac{x}{20}

The unknown is on top, so multiply both sides by

20

x = 20\sin 35^\circ

Evaluate with a calculator in degree mode

x \approx 11.5

Answer: $x \approx 11.5$

Example 2: unknown on top (tangent)

A wire is staked $30$ ft from the base of a pole and makes a $52^\circ$ angle with the ground. How tall is the pole, to the nearest tenth?

The pole is opposite the angle and

30

is adjacent — tangent

\tan 52^\circ = \dfrac{x}{30}

Multiply both sides by

30

x = 30\tan 52^\circ

Evaluate

x \approx 38.4

Answer: $x \approx 38.4$ ft

Example 3: unknown on the bottom

In a right triangle, the side opposite a $28^\circ$ angle measures $9$ . Find the hypotenuse $x$ , to the nearest tenth.

Opposite and hypotenuse means sine — the unknown is the bottom of the ratio

\sin 28^\circ = \dfrac{9}{x}

Solve for

x

by dividing

x = \dfrac{9}{\sin 28^\circ}

Evaluate

x \approx 19.2

Check: the hypotenuse should be the longest side, and

19.2 > 9

✓

Answer: $x \approx 19.2$

Try one yourself

Common questions

How do I decide between sine, cosine, and tangent?

Identify which two sides the problem involves — the one you know and the one you want. Then match the pair: opposite-hypotenuse is sine, adjacent-hypotenuse is cosine, opposite-adjacent is tangent. Only one ratio fits any given pair.

My calculator gives a strange answer. What went wrong?

Check that it is in degree mode, not radian mode. A giveaway: $\sin 35^\circ$ should be about $0.57$ ; if your calculator says something like $-0.43$ , it is in radians.

When do I multiply and when do I divide?

Look at where the unknown sits in the ratio. On top: multiply by the bottom number. On the bottom: swap it with the trig value, so the unknown equals the top number divided by the trig value.

Does it matter which acute angle I use?

You can use either, but the labels flip — the side opposite one acute angle is adjacent to the other. Use the angle the problem gives you and label everything from that angle's point of view.

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Solving for Missing Sides with Trig

Label the sides first

Pick the ratio that uses your two sides

Multiply or divide?

Worked examples

Example 1: unknown on top (sine)

Example 2: unknown on top (tangent)

Example 3: unknown on the bottom

Try one yourself

Common questions

Keep learning

Inverse Trig: Finding Missing Angles

Angles of Elevation & Depression

The Pythagorean Theorem & Its Converse

Special Right Triangles: 30-60-90