Chapter 6: Right Triangle Trigonometry

Side ratios, inverse trig setup, and the triangle laws used when right-triangle tools are not enough.

Chapter Overview

Right Triangle LadderUse sine with opposite and hypotenuse. Ramp RunUse cosine with adjacent and hypotenuse. Flagpole ShadowUse tangent with opposite and adjacent. Inverse AngleUse inverse trig to recover an angle. Non-Right TrianglePlaceholder for Law of Sines applications. SAS TrianglePlaceholder for Law of Cosines applications.

Trigonometry begins with a simple idea: an acute angle in a right triangle determines fixed ratios between side lengths. The names sine, cosine, and tangent are labels for those ratios.

  • Identify the hypotenuse first because it is always opposite the right angle.
  • Name the opposite and adjacent sides relative to the marked angle.
  • Choose sine, cosine, or tangent based on the two sides involved.
  • Use inverse trig functions to find an angle from a known ratio.
  • Use the Law of Sines and Law of Cosines when the triangle is not right.

6.1 SOHCAHTOA Ratios

SOHCAHTOA is a memory device for the three primary right-triangle ratios.

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$$ $$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$ $$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$$
  • $\theta$ is the reference angle.
  • The hypotenuse is the longest side and sits across from the right angle.
  • The opposite side sits across from $\theta$.
  • The adjacent side touches $\theta$ but is not the hypotenuse.

6.2 Right Triangle Setup

Before choosing a formula, label the triangle. Most trigonometry errors happen because the wrong side was called opposite or adjacent.

adjacent opposite hypotenuse theta

Worked Example 6.2.1 — Ladder Height

A ladder leans against a wall and makes a $62^\circ$ angle with the ground. The ladder is 18 feet long. How high up the wall does the ladder reach?

  • The ladder is the hypotenuse because it is across from the right angle.H=18
  • The wall height is opposite the angle.O=h
  • Use sine because sine connects opposite and hypotenuse.\sin 62^\circ=\frac{h}{18}
$$\sin 62^\circ=\frac{h}{18}$$ $$h=18\sin 62^\circ\approx 15.9$$

Worked Example 6.2.2 — Ramp Run

A 20-foot ramp makes a 12^\circ angle with the ground. Find the horizontal run.

  • The ramp is the hypotenuse.H=20
  • The horizontal run is adjacent to the angle.A=x
  • Use cosine.\cos 12^\circ=\frac{x}{20}
  • Solve.x=20\cos 12^\circ\approx 19.6

Worked Example 6.2.3 — Flagpole Shadow

A flagpole casts a 30-foot shadow when the angle of elevation to the sun is 38^\circ. Find the height.

  • The height is opposite the angle.O=h
  • The shadow is adjacent to the angle.A=30
  • Use tangent.\tan 38^\circ=\frac{h}{30}
  • Solve.h=30\tan 38^\circ\approx 23.4

6.3 Inverse Trigonometric Functions

Inverse trigonometric functions find an angle from a ratio. They should be used after the side relationship is known.

$$\theta=\sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)$$ $$\theta=\cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)$$ $$\theta=\tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$$

Worked Example 6.3.1 — Inverse Angle

A right triangle has opposite side 7 and hypotenuse 12 relative to angle \theta. Find \theta.

  • Use sine because opposite and hypotenuse are known.\sin\theta=\frac{7}{12}
  • Apply inverse sine.\theta=\sin^{-1}\left(\frac{7}{12}\right)
  • Evaluate.\theta\approx 35.7^\circ

6.4 Law of Sines

Use the Law of Sines when you know a matching side-angle pair in a non-right triangle.

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$

Worked Example 6.4.1 — Surveyor Triangle

In triangle ABC, A=42^\circ, B=65^\circ, and a=18. Find b.

  • Use the matching pair A and a.\frac{a}{\sin A}=\frac{b}{\sin B}
  • Substitute known values.\frac{18}{\sin42^\circ}=\frac{b}{\sin65^\circ}
  • Solve for b.b=\frac{18\sin65^\circ}{\sin42^\circ}\approx24.4

6.5 Law of Cosines

Use the Law of Cosines for SAS side finding or SSS angle finding. It generalizes the Pythagorean theorem by adding the angle correction term.

Side Finder Form (SAS)

$$c^2=a^2+b^2-2ab\cos C$$

Angle Finder Form (SSS)

$$C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)$$

Worked Example 6.5.1 — SAS Side Finder

Two sides of a triangle are 9 and 14, and the included angle is 50^\circ. Find the opposite side.

  • Use side-finder form.c^2=a^2+b^2-2ab\cos C
  • Substitute values.c^2=9^2+14^2-2(9)(14)\cos50^\circ
  • Take the square root.c\approx10.7

Worked Example 6.5.2 — SSS Angle Finder

A triangle has sides 7, 10, and 12. Find the angle opposite the side of length 12.

  • Use angle-finder form.C=\cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)
  • Substitute values.C=\cos^{-1}\left(\frac{7^2+10^2-12^2}{2(7)(10)}\right)
  • Evaluate.C\approx87.9^\circ