To solve a right triangle means to find all of its sides and angles. With one acute angle and one side, the ratios from Sine, Cosine and Tangent find the rest; the Pythagorean theorem and the angle sum help check.
- Mark the angle you know and label the sides from it: opposite, adjacent, hypotenuse.
- Pick the ratio that uses the side you know and the side you want.
- Write the equation, then solve for the unknown. Use a calculator in degree mode.
Idea clave
Label the sides from the known angle, choose the ratio that links the side you know to the side you want, write the equation, then solve. Finish with the angle sum ($90^\circ$ for the two acute angles) and the Pythagorean theorem as checks.
Por qué funciona
The equation always has the form $\text{ratio} = \frac{\text{one side}}{\text{another side}}$. If the unknown is on top, multiply. If the unknown is on the bottom, multiply both sides by it first, then divide. Writing the equation before reaching for the calculator keeps that straight.
Ejemplo resuelto: The unknown on top
A right triangle has a $36^\circ$ angle and a hypotenuse of $20$. Find the leg opposite the $36^\circ$ angle, to the nearest tenth.
- Opposite and hypotenuse: sine. $\sin 36^\circ = \frac{x}{20}$.
- $x = 20 \sin 36^\circ \approx 20(0.5878) \approx 11.8$.
Respuesta: $x \approx 11.8$
Ejemplo resuelto: The unknown on the bottom
A right triangle has a $25^\circ$ angle, and the leg opposite it is $8$. Find the hypotenuse, to the nearest tenth.
- Opposite and hypotenuse: $\sin 25^\circ = \frac{8}{x}$.
- Multiply by $x$: $x \sin 25^\circ = 8$. Divide: $x = \frac{8}{\sin 25^\circ} \approx \frac{8}{0.4226} \approx 18.9$.
- Check: the hypotenuse must be longer than the leg, and $18.9 > 8$.
Respuesta: $x \approx 18.9$
Ejemplo resuelto: Solving the whole triangle
In right $\triangle ABC$ (right angle at $C$), $m\angle A = 52^\circ$ and $AC = 10$. Find $m\angle B$, $BC$ and $AB$.
- $m\angle B = 90 - 52 = 38^\circ$.
- $AC$ is adjacent to $A$ and $BC$ is opposite: $\tan 52^\circ = \frac{BC}{10}$, so $BC = 10 \tan 52^\circ \approx 12.8$.
- $\cos 52^\circ = \frac{10}{AB}$, so $AB = \frac{10}{\cos 52^\circ} \approx 16.2$. Check: $10^2 + 12.8^2 \approx 263.8$ and $16.2^2 \approx 262.4$ (close, given rounding).
Respuesta: $m\angle B = 38^\circ$, $BC \approx 12.8$, $AB \approx 16.2$
Error común
Setting up $\sin 25^\circ = \frac{x}{8}$ when $x$ is the hypotenuse, or computing in radian mode. The hypotenuse always goes on the bottom of sine and cosine. If an answer looks strange (a leg longer than the hypotenuse, or $\sin 30^\circ$ coming out as $-0.988$), check the setup and make sure the calculator shows DEG.