Math Core

A right triangle has six parts: three sides and three angles. One of the angles is always $90^\circ$ , so if you know just two more parts, at least one of them a side, you can find everything else. Doing that is called solving the triangle, and the trig ratios are the tools.

Setting up an equation

Every trig ratio connects one angle to two sides. To find a missing side, pick the ratio that involves the angle you know, the side you know, and the side you want. Then solve the equation.

Label the sides as opposite, adjacent and hypotenuse from the known angle.
Choose sine, cosine or tangent (SOH-CAH-TOA) so that the equation contains exactly one unknown.
Solve, and use a calculator for the trig value.

Common mistake

Your calculator must be in degree mode for these problems. In radian mode, $\sin 35$ means the sine of $35$ radians, and every answer will be wrong. A quick test: $\sin 30$ should give exactly $0.5$ .

Worked example: The unknown is in the numerator

In right triangle $ABC$ , $\angle C = 90^\circ$ , $\angle A = 35^\circ$ and the hypotenuse $AB = 12$ . Find $BC$ and $AC$ to the nearest tenth.

Right triangle ABC with angle A = 35° and hypotenuse 12.

$BC = x$ is opposite the $35^\circ$ angle and we know the hypotenuse, so use sine:

\sin 35^\circ = \frac{x}{12} \quad\Longrightarrow\quad x = 12 \sin 35^\circ \approx 12(0.5736) \approx 6.9.

$AC = y$ is adjacent to the angle, so use cosine:

\cos 35^\circ = \frac{y}{12} \quad\Longrightarrow\quad y = 12 \cos 35^\circ \approx 9.8.

Check with the Pythagorean theorem: $6.88^2 + 9.83^2 \approx 47.4 + 96.6 = 144 = 12^2$ .

Tip

Round only at the very end. Keep the full calculator value (or type the whole expression $12\sin(35)$ at once) so rounding errors don't pile up.

When the unknown is in the denominator

Sometimes the side you want ends up on the bottom of the ratio. Multiply both sides by the unknown, then divide.

Worked example: Finding the hypotenuse

In right triangle $ABC$ , $\angle C = 90^\circ$ , $\angle A = 40^\circ$ and $BC = 7$ . Find the hypotenuse $AB$ to the nearest tenth.

Right triangle ABC with angle A = 40° and BC = 7.

$BC = 7$ is opposite the angle and $x$ is the hypotenuse, so

\sin 40^\circ = \frac{7}{x} \quad\Longrightarrow\quad x \sin 40^\circ = 7 \quad\Longrightarrow\quad x = \frac{7}{\sin 40^\circ} \approx \frac{7}{0.6428} \approx 10.9.

The hypotenuse must be longer than either leg, and $10.9 > 7$ , so the answer is reasonable. You could also use cosecant directly: $x = 7 \csc 40^\circ$ .

Finding an angle

To find an angle, you need to run a ratio backwards: given the value of $\tan A$ , what is $A$ ? That is what the inverse trig functions do. They are written $\sin^{-1}$ , $\cos^{-1}$ and $\tan^{-1}$ (the $-1$ means "inverse," not a reciprocal). You will study them in depth in a later unit. For now, think of them as calculator keys that turn a ratio into an angle.

Finding an acute angle from two sides

If you know two sides of a right triangle, write the ratio they form for the angle $\theta$ , then apply the inverse:

\sin\theta = \frac{\text{opp}}{\text{hyp}} \Rightarrow \theta = \sin^{-1}\!\left(\frac{\text{opp}}{\text{hyp}}\right), \quad \cos\theta = \frac{\text{adj}}{\text{hyp}} \Rightarrow \theta = \cos^{-1}\!\left(\frac{\text{adj}}{\text{hyp}}\right), \quad \tan\theta = \frac{\text{opp}}{\text{adj}} \Rightarrow \theta = \tan^{-1}\!\left(\frac{\text{opp}}{\text{adj}}\right)

Worked example: Two legs given

In right triangle $ABC$ , $\angle C = 90^\circ$ , $AC = 8$ and $BC = 5$ . Find $\angle A$ and $\angle B$ to the nearest tenth of a degree.

Right triangle ABC with legs 8 and 5.

From $A$ , the opposite side is $5$ and the adjacent side is $8$ , so

\tan A = \frac{5}{8} = 0.625 \quad\Longrightarrow\quad A = \tan^{-1}(0.625) \approx 32.0^\circ.

The acute angles are complementary, so $B \approx 90^\circ - 32.0^\circ = 58.0^\circ$ . The hypotenuse, if you need it, is $\sqrt{8^2 + 5^2} = \sqrt{89} \approx 9.4$ .

Solving the whole triangle

A complete solution lists all three sides and all three angles. Use given values whenever possible rather than values you've already rounded.

Worked example: One angle and one leg

Solve right triangle $ABC$ with $\angle C = 90^\circ$ , $\angle A = 52^\circ$ and $AC = 15$ . Round sides to the nearest tenth.

Right triangle ABC with angle A = 52° and AC = 15.

Angle $B$ . $B = 90^\circ - 52^\circ = 38^\circ$ .

Side $BC$ . It is opposite $A$ , and $AC$ is adjacent, so use tangent:

BC = 15 \tan 52^\circ \approx 19.2.

Side $AB$ . It is the hypotenuse, and $AC$ is adjacent, so use cosine:

\cos 52^\circ = \frac{15}{AB} \quad\Longrightarrow\quad AB = \frac{15}{\cos 52^\circ} \approx 24.4.

Summary. $A = 52^\circ$ , $B = 38^\circ$ , $C = 90^\circ$ , $AC = 15$ , $BC \approx 19.2$ , $AB \approx 24.4$ .

Check: the longest side is across from the largest angle, and $19.2^2 + 15^2 \approx 593.6$ , close to $24.4^2 \approx 595.4$ (the small gap is rounding).

Tip

Before you finish, check that the pieces fit: the angles add to $180^\circ$ , the hypotenuse is the longest side, and the larger acute angle is across from the longer leg.

Practice

Round side lengths to the nearest tenth and angles to the nearest tenth of a degree.

Practice 1

In the triangle below, find $BC$ .

Right triangle ABC with angle A = 41° and hypotenuse 20.