Chapter 8: Trigonometry: SOHCAHTOA
Learning Objectives
- Label the sides of a right triangle (opposite, adjacent, hypotenuse) relative to a given angle
- Define and apply the three primary trigonometric ratios: sine, cosine, tangent (SOH-CAH-TOA)
- Use trig ratios and inverse trig functions to find missing sides and angles in right triangles
- Solve angle-of-elevation and angle-of-depression problems
- Apply the Law of Sines and Law of Cosines to non-right triangles
- Choose the appropriate trig tool (SOH-CAH-TOA, Law of Sines, Law of Cosines) for a given triangle
8.1 The Three Trig Ratios: SOH-CAH-TOA
Trigonometry connects the angles of a right triangle to the ratios of its sides. For any acute angle $\theta$ in a right triangle, we name the three sides relative to that angle:
- Opposite — the side directly across from $\theta$ (not touching $\theta$)
- Adjacent — the leg that forms one side of $\theta$ (the non-hypotenuse leg touching $\theta$)
- Hypotenuse — the longest side, always opposite the right angle
Definition: The Three Trigonometric Ratios
For an acute angle $\theta$ in a right triangle:
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$Mnemonic — SOH-CAH-TOA: Some Old Horses Can Always Hear Their Owner's Approach.
The three reciprocal functions are defined as inverses of the primary ratios: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$. These appear more often in precalculus and calculus.
Right triangle with labeled sides — opposite, adjacent, and hypotenuse relative to angle θ at vertex A.
Figure 8.1 — Right Triangle: Sides Labeled Relative to Angle θ
Example 8.1 — Finding All Three Trig Ratios
In right $\triangle ABC$ with right angle at $C$, $AB = 13$, $BC = 5$, $AC = 12$, and $\angle A = \theta$. Find $\sin\theta$, $\cos\theta$, and $\tan\theta$.
Identify sides relative to $\angle A$:
- Opposite $\angle A$: side $BC = 5$
- Adjacent to $\angle A$: side $AC = 12$
- Hypotenuse: $AB = 13$
$$\sin\theta = \frac{5}{13} \qquad \cos\theta = \frac{12}{13} \qquad \tan\theta = \frac{5}{12}$$
Verify: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ ✓ (This is the 5-12-13 Pythagorean triple.)
Example 8.2 — Finding Two Ratios from One
In a right triangle, $\sin\theta = \dfrac{3}{5}$. Find $\cos\theta$ and $\tan\theta$.
From $\sin\theta = \frac{3}{5}$: opposite $= 3$, hypotenuse $= 5$.
By the Pythagorean Theorem: $\text{adj} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.
$$\cos\theta = \frac{4}{5} \qquad \tan\theta = \frac{3}{4}$$
Note: This is the 3-4-5 Pythagorean triple scaled by 1.
In a right triangle, the hypotenuse is 17 and one leg is 8. Find all three trig ratios for the angle $\theta$ that is opposite the leg of length 8.
Show Answer
Relative to $\theta$: opp $= 8$, adj $= 15$, hyp $= 17$.
$$\sin\theta = \frac{8}{17}, \quad \cos\theta = \frac{15}{17}, \quad \tan\theta = \frac{8}{15}$$ (This is the 8-15-17 Pythagorean triple.)
8.2 Using Trig Ratios to Solve Triangles
Once you write a trig ratio as an equation, you can solve it algebraically for any unknown side. Most geometry problems use degree mode on a calculator.
Solving Right Triangle Summary
- To find a missing side: Write the trig ratio, then solve for the unknown.
Example: $\sin\theta = \frac{\text{opp}}{\text{hyp}}$ → if opp is unknown: $\text{opp} = \text{hyp} \cdot \sin\theta$ - To find a missing angle: Use the inverse trig function.
If $\sin\theta = r$, then $\theta = \sin^{-1}(r)$. Similarly for $\cos^{-1}$ and $\tan^{-1}$. - Angle sum check: The two acute angles of a right triangle always sum to $90°$.
Example 8.3 — Finding a Missing Side
In a right triangle, an acute angle is $35°$ and the hypotenuse is $20$. Find the leg opposite the $35°$ angle.
$$\sin 35° = \frac{x}{20} \implies x = 20\sin 35° \approx 20(0.5736) \approx \mathbf{11.47}$$
Example 8.4 — Finding Missing Angles
In a right triangle, the two legs measure 6 and 8. Find both acute angles.
Let $\theta$ be opposite the leg of length 6. Then $\tan\theta = \dfrac{6}{8} = 0.75$.
$$\theta = \tan^{-1}(0.75) \approx \mathbf{36.87°}$$
The other acute angle $= 90° - 36.87° = \mathbf{53.13°}$.
Check: $36.87° + 53.13° = 90°$ ✓. (This is the 6-8-10 right triangle — a 3-4-5 triple scaled by 2.)
Example 8.5 — Real-World Application
A ramp rises 4 feet vertically over a horizontal distance of 30 feet. Find the angle of inclination.
$$\tan\theta = \frac{4}{30} = \frac{2}{15} \implies \theta = \tan^{-1}\!\left(\frac{2}{15}\right) \approx \mathbf{7.6°}$$
A building casts a shadow of 40 m when the sun is at an elevation angle of 55°. How tall is the building?
Show Answer
$$\tan 55° = \frac{h}{40} \implies h = 40\tan 55° \approx 40(1.4281) \approx \mathbf{57.1 \text{ m}}$$
AP/SAT Tip: Always draw a diagram first. Label the known sides, identify which angle you are working from, then write the opp/adj/hyp labels relative to that angle — not the other one. A common error is labeling opp and adj for the wrong angle.
Right triangle with a 30° angle — observe the relationship between the angle and the side lengths.
Figure 8.2 — Right Triangle with 30° Angle: Sides and Ratios
8.3 Angles of Elevation and Depression
Angle of Elevation vs. Angle of Depression
- Angle of elevation: The angle measured upward from a horizontal line to the line of sight toward an object above.
- Angle of depression: The angle measured downward from a horizontal line to the line of sight toward an object below.
- Key relationship: The angle of elevation from point $A$ to point $B$ equals the angle of depression from $B$ to $A$. These are alternate interior angles formed by the horizontal lines and the line of sight (a transversal).
In both cases, the horizontal distance and the vertical height form the two legs of a right triangle, and the line of sight is the hypotenuse. Use SOH-CAH-TOA (almost always tangent, since the right angle is between the horizontal and vertical) to solve.
Example 8.6 — Angle of Elevation
From a point 60 m from the base of a tree, the angle of elevation to the top is 40°. Find the height of the tree.
The horizontal distance (adjacent) $= 60$ m; height (opposite) $= h$; angle $= 40°$.
$$\tan 40° = \frac{h}{60} \implies h = 60\tan 40° \approx 60(0.8391) \approx \mathbf{50.35 \text{ m}}$$
Example 8.7 — Angle of Depression
From the top of a 150-ft cliff, the angle of depression to a boat is 25°. Find the horizontal distance from the base of the cliff to the boat.
The angle of depression $= 25°$. The cliff height (opposite) $= 150$ ft; horizontal distance (adjacent) $= d$.
$$\tan 25° = \frac{150}{d} \implies d = \frac{150}{\tan 25°} \approx \frac{150}{0.4663} \approx \mathbf{321.7 \text{ ft}}$$
A person with eyes at 5.5 ft above the ground sees a ball on the ground at a 10° angle of depression. What is the horizontal distance from the person's position to the ball?
Show Answer
$$\tan 10° = \frac{5.5}{d} \implies d = \frac{5.5}{\tan 10°} \approx \frac{5.5}{0.1763} \approx \mathbf{31.2 \text{ ft}}$$
AP Tip: In word problems involving elevation or depression, always sketch the triangle first. Mark the horizontal line, draw the line of sight, and label the angle. This immediately shows you which sides are opp, adj, and hyp — and prevents the most common setup errors.
8.4 Law of Sines and Law of Cosines (Optional — Non-Right Triangles)
SOH-CAH-TOA only works for right triangles. When a triangle has no right angle, we need two more powerful tools: the Law of Sines and the Law of Cosines.
Convention: in $\triangle ABC$, side $a$ is opposite $\angle A$, side $b$ is opposite $\angle B$, and side $c$ is opposite $\angle C$.
Law of Sines
For any triangle $\triangle ABC$:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$Use when given: AAS (two angles and a non-included side) or ASA (two angles and the included side) — and also SSA, the ambiguous case (may produce 0, 1, or 2 solutions).
Law of Cosines
For any triangle $\triangle ABC$:
$$c^2 = a^2 + b^2 - 2ab\cos C$$(and symmetrically for $a^2$ and $b^2$). This generalizes the Pythagorean Theorem: when $C = 90°$, $\cos 90° = 0$ and we recover $c^2 = a^2 + b^2$.
Use when given: SAS (two sides and the included angle) or SSS (all three sides).
| Known Information | Method | Notes |
|---|---|---|
| Two angles + any side (AAS or ASA) | Law of Sines | Find third angle first (angle sum = 180°) |
| Two sides + angle opposite one of them (SSA) | Law of Sines | Ambiguous case — check for 0, 1, or 2 solutions |
| Two sides + included angle (SAS) | Law of Cosines | Solve for the third side directly |
| All three sides (SSS) | Law of Cosines | Rearrange to find any angle: $\cos C = \frac{a^2+b^2-c^2}{2ab}$ |
Example 8.8 — Law of Sines
In $\triangle ABC$, $\angle A = 40°$, $\angle B = 70°$, and $a = 15$. Find side $b$.
First: $\angle C = 180° - 40° - 70° = 70°$. Then apply the Law of Sines:
$$\frac{15}{\sin 40°} = \frac{b}{\sin 70°} \implies b = \frac{15\sin 70°}{\sin 40°} \approx \frac{15(0.9397)}{0.6428} \approx \mathbf{21.93}$$
Example 8.9 — Law of Cosines
In $\triangle ABC$, $a = 8$, $b = 11$, and $\angle C = 60°$. Find side $c$.
$$c^2 = 8^2 + 11^2 - 2(8)(11)\cos 60° = 64 + 121 - 176\left(\frac{1}{2}\right) = 185 - 88 = 97$$
$$c = \sqrt{97} \approx \mathbf{9.85}$$
In $\triangle PQR$, $\angle P = 50°$, $\angle Q = 65°$, and $p = 20$. Find side $q$.
Show Answer
$$\frac{20}{\sin 50°} = \frac{q}{\sin 65°} \implies q = \frac{20\sin 65°}{\sin 50°} \approx \frac{20(0.9063)}{0.7660} \approx \mathbf{23.66}$$
General triangle (non-right) illustrating the Law of Cosines — sides a, b, c labeled opposite their respective angles.
Figure 8.3 — General Triangle: Law of Cosines Relationship
Practice Problems
In right $\triangle ABC$ with right angle at $C$, $BC = 7$ and $AC = 24$. Find $\sin A$, $\cos A$, $\tan A$, and $\sin B$.
Show Solution
Relative to $\angle A$: opp $= BC = 7$, adj $= AC = 24$, hyp $= 25$.
$\sin A = \dfrac{7}{25}$, $\cos A = \dfrac{24}{25}$, $\tan A = \dfrac{7}{24}$.
For $\angle B$: opp $= AC = 24$, so $\sin B = \dfrac{24}{25}$.
In a right triangle, one acute angle is $42°$ and the hypotenuse is $25$. Find the leg opposite the $42°$ angle.
Show Solution
A ladder 20 ft long makes a 65° angle with the ground. How high up the wall does the top of the ladder reach?
Show Solution
$$\sin 65° = \frac{h}{20} \implies h = 20\sin 65° \approx 20(0.9063) \approx \mathbf{18.13 \text{ ft}}$$
From the top of a 200-ft lighthouse, the angle of depression to a ship is 18°. Find the horizontal distance from the base of the lighthouse to the ship.
Show Solution
$$\tan 18° = \frac{200}{d} \implies d = \frac{200}{\tan 18°} \approx \frac{200}{0.3249} \approx \mathbf{615.7 \text{ ft}}$$
A triangle has sides $a = 9$, $b = 12$, and included angle $C = 110°$. Use the Law of Cosines to find side $c$.
Show Solution
In $\triangle ABC$, $\angle A = 35°$, $\angle B = 85°$, and $a = 14$. Find side $b$ using the Law of Sines.
Show Solution
$$\frac{14}{\sin 35°} = \frac{b}{\sin 85°} \implies b = \frac{14\sin 85°}{\sin 35°} \approx \frac{14(0.9962)}{0.5736} \approx \mathbf{24.31}$$
A surveyor stands 150 m from a hill. The angle of elevation to the hilltop is 28°. A cell tower sits on top of the hill, and the angle of elevation to the top of the tower is 35°. Find the height of the cell tower alone.
Show Solution
Height to top of tower: $h_2 = 150\tan 35° \approx 150(0.7002) \approx 105.03$ m.
Tower height $= h_2 - h_1 \approx 105.03 - 79.76 \approx \mathbf{25.27 \text{ m}}$.
AP Problem: Two ships leave a harbor simultaneously. Ship A travels N 40° E for 60 km; Ship B travels S 30° E for 80 km. Using the Law of Cosines, find the distance between the ships.
Show Solution
Apply the Law of Cosines with $a = 60$, $b = 80$, $C = 110°$:
$$c^2 = 60^2 + 80^2 - 2(60)(80)\cos 110°$$ $$= 3600 + 6400 - 9600(-0.3420) = 10000 + 3283.2 = 13283.2$$ $$c = \sqrt{13283.2} \approx \mathbf{115.3 \text{ km}}$$
📋 Chapter Summary
Laws for Oblique Triangles
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$. Use when given AAS, ASA, or SSA (check for ambiguous case).
$c^2 = a^2 + b^2 - 2ab\cos C$. Use when given SAS or SSS. Reduces to Pythagorean theorem when $C = 90°$.
$A = \frac{1}{2}ab\sin C$ — area of triangle given two sides and included angle.
Given $a$, $b$, $A$: may have 0, 1, or 2 triangles. Check if $a \geq b$ (one solution) or $a < b\sin A$ (no solution) or between (two solutions).
When to Use Each Law
- Right triangle → SOH-CAH-TOA or Pythagorean theorem
- AAS or ASA → Law of Sines (angle-angle-side or angle-side-angle)
- SSA → Law of Sines, but check for ambiguous case
- SAS → Law of Cosines (two sides and included angle)
- SSS → Law of Cosines (all three sides known)