Lesson 117: Using Trigonometric Ratios

New Concept

In any right triangle, there are six trigonometric ratios that can be written using two side lengths of the triangle in relation to the angles.

What’s next

Next, you’ll learn the key ratios (SOH-CAH-TOA) and apply them to find missing side lengths and angles in right triangles.

Property

To remember the three main trigonometric ratios, use the mnemonic SOH-CAH-TOA.

Explanation

This is a super-helpful memory trick! SOH-CAH-TOA stands for Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, and Tangent is Opposite over Adjacent. It's the key to matching the right trig ratio with the correct sides of a right triangle, making homework feel like less of a puzzle and more of a win.

Examples

• In a right triangle with angle B, opposite leg = 5, adjacent leg = 12, and hypotenuse = 13:
  $$\operatorname{sin} B = \frac{5}{13} \text{ and } \operatorname{cos} B = \frac{12}{13}$$
  .
• For angle A in a right triangle with opposite leg = 20 and adjacent leg = 21, the tangent is:
  $$\operatorname{tan} A = \frac{20}{21}$$
  .
• Given a right triangle with angle A, opposite leg = 8, and hypotenuse = 17, the sine is:
  $$\operatorname{sin} A = \frac{8}{17}$$
  .

Let's find all six trigonometric ratios for an angle, even when one side length is missing. This example uses the Pythagorean Theorem, a fundamental tool for right triangles.

Example Problem

Find all six trigonometric ratios for $\angle A$ in a right triangle where the adjacent leg $b=15$ and the hypotenuse $c=17$.

Property

Cosecant (csc), secant (sec), and cotangent (cot) are the reciprocal ratios of sine, cosine, and tangent.

Explanation

Think of these as the "upside-down" versions of SOH-CAH-TOA. Cosecant flips sine, secant flips cosine, and cotangent flips tangent. If you know the first three ratios, just flip the fraction to find these! It’s like a buy-three-get-three-free deal for trigonometric ratios, which makes your life six times easier when solving problems.

Examples

• If $\operatorname{sin} A = \frac{3}{5}$ and $\operatorname{cos} A = \frac{4}{5}$, then their reciprocals are:
  $$\operatorname{csc} A = \frac{5}{3} \text{ and } \operatorname{sec} A = \frac{5}{4}$$
  .
• If $\operatorname{tan} A = \frac{3}{4}$, then the cotangent is the reciprocal:
  $$\operatorname{cot} A = \frac{4}{3}$$
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• In a triangle where $\operatorname{sin} B = \frac{5}{13}$, the cosecant is:
  $$\operatorname{csc} B = \frac{13}{5}$$
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With just one angle and one side, we can unlock the other sides of the triangle. This example applies the trigonometric ratios to find a missing side length.

Example Problem

Find the value of $x$ in a right triangle with a $34^{\circ}$ angle, opposite side $x$, and hypotenuse 15. Round to the nearest hundredth.

Property

Use a known angle and a known side length to find an unknown side length by setting up a trigonometric equation.

Explanation

Got an angle and one side length but need another? Trig ratios to the rescue! Pick the right ratio (SOH-CAH-TOA) that connects your known angle, known side, and the side you're looking for. Then, you just solve the simple equation for the missing length. It’s like being a super-sleuth for triangle mysteries!

Examples

• To find side $x$ opposite a $28^{\circ}$ angle with an adjacent side of 9:
  $$\operatorname{tan} 28^{\circ} = \frac{x}{9} \rightarrow x = 9 \cdot \operatorname{tan} 28^{\circ} \approx 4.79$$
  .
• To find side $y$ opposite a $52^{\circ}$ angle with a hypotenuse of 12:
  $$\operatorname{sin} 52^{\circ} = \frac{y}{12} \rightarrow y = 12 \cdot \operatorname{sin} 52^{\circ} \approx 9.46$$
  .
• To find side $x$ adjacent to a $52^{\circ}$ angle with a hypotenuse of 12:
  $$\operatorname{cos} 52^{\circ} = \frac{x}{12} \rightarrow x = 12 \cdot \operatorname{cos} 52^{\circ} \approx 7.39$$
  .