Honors Geometry Unit 8 Review: Right Triangles and Trigonometry

Special right triangles have fixed angles, so their sides always follow the same ratio. That lets you find exact side lengths without a calculator.

In a 45-45-90 triangle the legs are equal and the hypotenuse is a leg times $\sqrt{2}$, so the ratio is $x : x : x\sqrt{2}$. You get one whenever you cut a square along its diagonal, so a square with side 6 has a diagonal of $6\sqrt{2}$.

In a 30-60-90 triangle the ratio is $x : x\sqrt{3} : 2x$. The short leg $x$ is opposite 30°, the long leg $x\sqrt{3}$ is opposite 60°, and the hypotenuse $2x$ is opposite 90°. Drawing an altitude in an equilateral triangle splits it into two of these.

The reliable method is to find $x$ first, then build the other sides from it.

Worked example. A 30-60-90 triangle has a long leg of 9. Set $x\sqrt{3} = 9$, so $x = \frac{9}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}$. The hypotenuse is $2x = 6\sqrt{3}$. Check with the Pythagorean Theorem. $(3\sqrt{3})^2 + 9^2 = 27 + 81 = 108$, and $(6\sqrt{3})^2 = 36 \cdot 3 = 108$, so the sides work.

Pick one acute angle $\theta$ in a right triangle. The side across from it is the opposite side, the leg touching it is the adjacent side, and the side across from the right angle is the hypotenuse. SOH-CAH-TOA gives the three ratios. $\sin\theta = \frac{\text{opp}}{\text{hyp}}$, $\cos\theta = \frac{\text{adj}}{\text{hyp}}$, and $\tan\theta = \frac{\text{opp}}{\text{adj}}$.

These ratios depend only on the angle, because any two right triangles with the same acute angle are similar. To find a side, choose the ratio that uses the side you know and the side you want. To find an angle from two sides, use an inverse function like $\sin^{-1}$. Keep your calculator in degree mode.

To fully solve the triangle, remember that the two acute angles add to 90°.

Worked example. On level ground, you measure an angle of elevation of 35° from ground level to the top of a vertical tree. The measurement point is 40 ft from the tree's base. The height is opposite the angle and 40 ft is adjacent, so use tangent. $\tan 35° = \frac{h}{40}$, so $h = 40\tan 35° \approx 40(0.700) \approx 28.0$ ft. For an angle, if the opposite side is 5 and the hypotenuse is 13, then $\theta = \sin^{-1}\left(\frac{5}{13}\right) \approx 22.6°$, and the other acute angle is about $67.4°$.

In any right triangle, $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs and $c$ is the hypotenuse. To find the hypotenuse, add the squares of the legs and take the square root. To find a leg, subtract the square of the known leg from the square of the hypotenuse, then take the square root. With a hypotenuse of 10 and a leg of 6, the other leg is $\sqrt{100 - 36} = 8$.

The converse runs the other way. If three side lengths satisfy $a^2 + b^2 = c^2$ with $c$ as the longest side, the triangle is a right triangle. Honors courses usually add the inequality versions, which let you classify any triangle from its sides alone.

The theorem also gives you the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, and the space diagonal of a box, $d = \sqrt{l^2 + w^2 + h^2}$.

Worked example. Classify triangles with sides 8, 15, 17 and 6, 7, 10. For the first, $8^2 + 15^2 = 64 + 225 = 289 = 17^2$, so it's a right triangle. For the second, $6^2 + 7^2 = 36 + 49 = 85$ and $10^2 = 100$. Since $85 < 100$, the triangle is obtuse, with the obtuse angle opposite the side of length 10.

An oblique triangle has no right angle, so SOH-CAH-TOA doesn't apply directly. Two laws work for any triangle, where side $a$ is opposite angle $A$, and so on.

The Law of Sines says $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use it for ASA, AAS, or SSA by matching a known side with its opposite angle. In ASA, find the third angle first to get that pair. SSA is the ambiguous case. Since $\sin\theta = \sin(180° - \theta)$, an angle found with $\sin^{-1}$ might have an obtuse partner, so the given information can produce zero, one, or two triangles.

The Law of Cosines says $c^2 = a^2 + b^2 - 2ab\cos C$, where $C$ is the angle between sides $a$ and $b$. Use it for SAS or SSS. To find an angle from three sides, use $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Inverse cosine returns angles up to 180°, so it has no ambiguous case.

Worked example. Given $a = 7$, $b = 9$, and $C = 120°$ (SAS), find $c$. $c^2 = 49 + 81 - 2(7)(9)\cos 120° = 130 - 126(-0.5) = 130 + 63 = 193$, so $c = \sqrt{193} \approx 13.89$. Now you have the pair $c$ and $C$, so you can switch to the Law of Sines for the remaining angles.