8.3 Trigonometric ratios and solving right triangles

Trigonometric ratios connect the angles of a right triangle to the lengths of its sides. By defining sine, cosine, and tangent, you gain tools to find any missing side or angle in a right triangle, as long as you know at least one side and one acute angle (or two sides). This section covers how to set up and use these ratios, how to apply inverse trig functions to find angles, and how trig connects back to the Pythagorean Theorem.

Trigonometric Ratios

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Sine, cosine, and tangent ratios

Every right triangle has two acute angles. When you pick one of those angles (call it $\theta$), the three sides get specific names relative to that angle:

• Opposite: the side directly across from $\theta$
• Adjacent: the side next to $\theta$ (that isn't the hypotenuse)
• Hypotenuse: the longest side, always across from the right angle

The three trig ratios are defined as follows:

• Sine: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
  • Example: If the opposite side is 3 and the hypotenuse is 5, then $\sin \theta = \frac{3}{5}$
• Cosine: $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
  • Example: If the adjacent side is 4 and the hypotenuse is 5, then $\cos \theta = \frac{4}{5}$
• Tangent: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
  • Example: If the opposite side is 3 and the adjacent side is 4, then $\tan \theta = \frac{3}{4}$

All three examples above come from the same 3-4-5 right triangle. That's a good way to check your understanding: the ratios should be consistent with each other. Also notice that tangent can be derived from the other two: $\tan \theta = \frac{\sin \theta}{\cos \theta}$, since dividing $\frac{\text{opp}/\text{hyp}}{\text{adj}/\text{hyp}}$ cancels the hypotenuse.

The mnemonic SOH-CAH-TOA keeps these straight:

• Sine = Opposite / Hypotenuse
• Cosine = Adjacent / Hypotenuse
• Tangent = Opposite / Adjacent

A common mistake is mixing up which side is "opposite" and which is "adjacent." These labels change depending on which acute angle you're looking at. Always identify your angle first, then label the sides relative to it. If you switch to the other acute angle, the opposite and adjacent sides swap, but the hypotenuse stays the same.

Solving right triangles

"Solving" a right triangle means finding all unknown sides and angles. The approach depends on what information you start with.

When you're given an angle and one side:

1. Label the sides as opposite, adjacent, or hypotenuse relative to the known angle.
2. Pick the trig ratio that uses the known side and the unknown side.
3. Substitute and solve.

Example: In a right triangle, $\theta = 30°$ and the hypotenuse is 10. Find the side opposite $\theta$.

1. You know the hypotenuse and want the opposite side, so use sine.
2. Set up the equation: $\sin 30° = \frac{\text{opposite}}{10}$
3. Solve: $\text{opposite} = 10 \cdot \sin 30° = 10 \cdot 0.5 = 5$

When you're given two sides and need an angle:

1. Label the two known sides relative to the unknown angle.
2. Pick the trig ratio that involves both known sides.
3. Use the inverse trig function to solve for the angle.

Example: The opposite side is 5 and the hypotenuse is 13. Find angle $\theta$.

1. You have opposite and hypotenuse, so use sine.
2. Set up the equation: $\sin \theta = \frac{5}{13}$
3. Apply the inverse: $\theta = \sin^{-1}\left(\frac{5}{13}\right) \approx 22.6°$

On your calculator, $\sin^{-1}$ (also written $\arcsin$) "undoes" the sine function to return the angle. The same logic applies with $\cos^{-1}$ and $\tan^{-1}$. Make sure your calculator is set to degree mode, not radians. If you're getting weird answers (like 0.3938 instead of 22.6°), that's almost certainly a radians issue.

Fully solving the triangle: Once you find one missing piece, keep going. If you found one acute angle, the other is $90° - \theta$ (since the angles in a triangle sum to 180° and one angle is already 90°). Then use the Pythagorean Theorem or another trig ratio to get any remaining sides.

Real-world applications of trigonometry

Trig ratios show up whenever a real-world situation can be modeled with a right triangle. Two terms you'll see often in word problems:

• Angle of elevation: the angle measured upward from the horizontal (you're looking up at something)
• Angle of depression: the angle measured downward from the horizontal (you're looking down at something)

Both create right triangles you can solve with trig.

Steps for word problems:

1. Sketch the situation and identify the right triangle.
2. Label the known sides and angles, and mark what you need to find.
3. Choose the appropriate trig ratio.
4. Set up the equation, substitute, and solve.
5. Interpret your answer in context (include units, round appropriately).

Example: A 5 m ladder leans against a wall, forming a 70° angle with the ground. How high up the wall does the ladder reach?

1. The ladder is the hypotenuse (5 m). The wall height is the side opposite the 70° angle. The ground is the adjacent side.
2. You know the hypotenuse and want the opposite side, so use sine.
3. $\sin 70° = \frac{\text{wall height}}{5}$
4. $\text{wall height} = 5 \cdot \sin 70° \approx 5 \cdot 0.9397 \approx 4.7 \text{ m}$
5. The ladder reaches approximately 4.7 m up the wall.

Trigonometry vs. Pythagorean Theorem

Both tools work with right triangles, but they solve different types of problems:

• The Pythagorean Theorem ($a^2 + b^2 = c^2$) relates the three sides to each other. Use it when you know two sides and need the third, with no angle information required.
  • Example: Sides of 3 and 4 give a hypotenuse of $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$
• Trig ratios relate sides to angles. Use them when an angle is involved, either as a given or as the unknown.

Quick decision guide: if the problem gives you or asks for an angle measure, reach for trig. If it only involves side lengths, the Pythagorean Theorem is likely your move.

A useful connection between the two is the Pythagorean identity:

$\sin^2 \theta + \cos^2 \theta = 1$

This follows directly from the Pythagorean Theorem. Since $\sin \theta = \frac{\text{opp}}{\text{hyp}}$ and $\cos \theta = \frac{\text{adj}}{\text{hyp}}$, squaring and adding gives:

$\frac{\text{opp}^2}{\text{hyp}^2} + \frac{\text{adj}^2}{\text{hyp}^2} = \frac{\text{opp}^2 + \text{adj}^2}{\text{hyp}^2} = \frac{\text{hyp}^2}{\text{hyp}^2} = 1$

You can verify this with any angle. For $\theta = 30°$:

$\sin^2 30° + \cos^2 30° = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$

This identity is worth memorizing. It shows up again in later math courses and can help you check your work: if your sine and cosine values for the same angle don't satisfy this equation, something went wrong.