Prerequisites

Right Triangle Trigonometry

AP PRECALCULUS — PREREQUISITE REVIEW

Right Triangle Trigonometry

Notes — Prerequisite Topic 3

💡 Learning Objectives

By the end of this lesson you will be able to:

  • Identify the opposite side, adjacent side, and hypotenuse relative to a given acute angle
  • Use SOH-CAH-TOA to write the six trigonometric ratios
  • Evaluate the trig ratios of 30°, 45°, and 60° without a calculator
  • Solve right triangles: find unknown sides and angles from given information
  • Apply right triangle trig to angle-of-elevation and angle-of-depression problems
  • Recognize the Pythagorean theorem as a consistency check for computed sides

1. Parts of a Right Triangle

Every right triangle contains a 90° angle and two acute angles. The names we give the sides depend on which acute angle we are working with.

Standard labelling relative to angle θ. The hypotenuse is always the longest side (across from the right angle). The opposite side is across from θ; the adjacent side is next to θ and not the hypotenuse.

⚠️ Opposite and adjacent switch when you switch angles

In the same triangle, the side that is ‘opposite’ from the 30° angle is ‘adjacent’ to the 60° angle. Always re-identify the three sides relative to the angle you are actually using.

2. The Six Trigonometric Ratios

For an acute angle θ in a right triangle, the three primary ratios are built from pairs of side lengths.

The mnemonic SOH-CAH-TOA encodes the three primary trigonometric ratios. The three reciprocal ratios are built from the same three pairs.

Ratio

Definition

Reciprocal

sin θ

opposite / hypotenuse

csc θ = 1 / sin θ

cos θ

adjacent / hypotenuse

sec θ = 1 / cos θ

tan θ

opposite / adjacent

cot θ = 1 / tan θ

💡 Quotient identity

The tangent is the ratio of the sine to the cosine:

tan θ = sin θ / cos θ.

This follows immediately from the definitions — divide opposite/hypotenuse by adjacent/hypotenuse and the hypotenuse cancels.

3. The Pythagorean Theorem

In any right triangle with legs a and b and hypotenuse c:

a² + b² = c².

This identity is the backbone of every side-length computation in right-triangle trigonometry. Given any two sides of a right triangle, you can always find the third.

📘 Example — Finding a missing side

A right triangle has legs of length 7 and 24. Find the hypotenuse.

c² = 7² + 24² = 49 + 576 = 625.

c = √625 = 25.

4. Special Right Triangles

Two right triangles appear so often that their exact side ratios are worth memorizing.

Left: the 45°-45°-90° triangle has legs of equal length and hypotenuse √2 times a leg. Right: the 30°-60°-90° triangle has sides in the ratio 1 : √3 : 2 (opposite 30°, opposite 60°, hypotenuse).

Angle θ

sin θ

cos θ

tan θ

30°

1/2

√3/2

√3/3

45°

√2/2

√2/2

1

60°

√3/2

1/2

√3

📘 Example — Using a special triangle

In a 30°-60°-90° triangle, the hypotenuse measures 10. Find the lengths of the two legs.

The side opposite 30° is half the hypotenuse: 5.

The side opposite 60° is (√3/2) · 10 = 5√3.

Check: 5² + (5√3)² = 25 + 75 = 100 = 10². ✓

5. Solving Right Triangles

To solve a right triangle means to find every unknown side and every unknown angle. Start by identifying what is given and choose a trig ratio whose definition connects what you know to what you want.

📘 Example — Given an angle and a side

In a right triangle, one acute angle is 37° and its opposite side is 15. Find the hypotenuse.

sin 37° = opposite / hypotenuse = 15 / c, so c = 15 / sin 37° ≈ 15 / 0.6018 ≈ 24.92.

📘 Example — Given two sides

A right triangle has legs 8 and 15 and hypotenuse 17. Find the acute angle opposite the side of length 8.

tan θ = 8 / 15 ≈ 0.5333, so θ = arctan(0.5333) ≈ 28.07°.

6. Angles of Elevation and Depression

💡 Two perspectives, one calculation

  • Angle of elevation: measured upward from the horizontal to the object above.
  • Angle of depression: measured downward from the horizontal to the object below.

From the viewer's horizontal, these angles are always measured off the horizontal line — not the vertical.

📘 Example — Angle of elevation

From a point on level ground 50 m from the base of a tower, the angle of elevation to the top of the tower is 62°. How tall is the tower?

tan 62° = height / 50, so height = 50 · tan 62° ≈ 50 · 1.8807 ≈ 94.03 m.

7. Summary

  • Label the three sides relative to the angle of interest before writing any ratio
  • Remember SOH-CAH-TOA; the reciprocals are csc, sec, cot
  • Use tan θ = sin θ / cos θ and a² + b² = c² as consistency checks
  • Memorize the exact ratios for 30°, 45°, and 60°
  • For word problems, sketch the triangle first, mark the given information, and choose the ratio that connects knowns to unknowns

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