Right-Triangle Ratios Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice right-triangle ratios and right triangle trigonometry: sine, cosine, and tangent using SOHCAHTOA, the reciprocal trig ratios (secant, cosecant, cotangent), finding missing sides with the Pythagorean theorem, special angle values for \(30^\circ\), \(45^\circ\), and \(60^\circ\) (including \(30\text{-}60\text{-}90\) and \(45\text{-}45\text{-}90\) triangles), and complementary angle identities like \(\sin(90^\circ-\theta)=\cos\theta\) and \(\tan(90^\circ-\theta)=\cot\theta\). If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
How this right-triangle ratios practice works
1. Take the quiz: answer the right-triangle trigonometric ratio questions at the top of the page.
2. Open the lesson (optional): review SOHCAHTOA, reciprocal trig ratios, special angles, and complementary identities with clear examples.
3. Retry: return to the quiz and apply right triangle ratio rules immediately.
What you will learn in the right-triangle ratios lesson
SOHCAHTOA foundations
Opposite, adjacent, hypotenuse (relative to the angle \(\theta\))
\(30\text{-}60\text{-}90\) and \(45\text{-}45\text{-}90\) triangle side ratios
Evaluate expressions like \(\csc(60^\circ)\), \(\tan(30^\circ)\), and \(\cos(45^\circ)\)
Solving right triangles
Pythagorean theorem to find missing sides
Use a trig ratio + one side to find another side (e.g., use \(\tan\theta\) to get opposite from adjacent)
Complementary angles (\(\theta\) and \(90^\circ-\theta\)) and cofunction identities
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing right-triangle ratios.
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Right-Triangle Ratios I
Step-by-Step Guide
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Right-Triangle Ratios I Lesson
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Lesson Overview
Lesson overview
Purpose: Build a clear understanding of right-triangle ratios (right triangle trigonometry) so you can use SOHCAHTOA to compute \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\), use reciprocal trig ratios (\(\sec\theta\), \(\csc\theta\), \(\cot\theta\)), apply the Pythagorean theorem to find missing sides, evaluate special angle values (\(30^\circ\), \(45^\circ\), \(60^\circ\)), and use complementary angle identities like \(\sin(90^\circ-\theta)=\cos\theta\).
Success criteria
Identify opposite, adjacent, and hypotenuse relative to a given acute angle \(\theta\).
Use SOHCAHTOA correctly: \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\).
Use reciprocal ratios: \(\sec\theta=\dfrac{1}{\cos\theta}\), \(\csc\theta=\dfrac{1}{\sin\theta}\), \(\cot\theta=\dfrac{1}{\tan\theta}\).
Find missing sides using the Pythagorean theorem: \(a^2+b^2=c^2\) (where \(c\) is the hypotenuse).
Evaluate trig ratios at special angles \(30^\circ\), \(45^\circ\), \(60^\circ\) using special triangles.
Use complementary angle identities in right triangles: \(\sin(90^\circ-\theta)=\cos\theta\), \(\cos(90^\circ-\theta)=\sin\theta\), \(\tan(90^\circ-\theta)=\cot\theta\).
Solve basic right triangle applications (e.g., missing side, angle of elevation).
Key vocabulary
Right triangle: a triangle with a \(90^\circ\) angle.
Hypotenuse: the side opposite the \(90^\circ\) angle (longest side).
Opposite / adjacent: defined relative to the chosen acute angle \(\theta\).
Trigonometric ratios: \(\sin\), \(\cos\), \(\tan\) as side-length ratios in right triangles.
Reciprocal ratios: \(\csc\), \(\sec\), \(\cot\) are reciprocals of \(\sin\), \(\cos\), \(\tan\).
Complementary angles: two angles that add to \(90^\circ\).
Quick pre-check
Pre-check 1: Which ratio defines \(\sin\theta\) in a right triangle?
Hint: SOH - sine equals opposite over hypotenuse.
Pre-check 2: In a right triangle, if the opposite side is \(6\) and the hypotenuse is \(10\), what is \(\sin\theta\)?
Hint: \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}=\dfrac{6}{10}\), then simplify.
SOHCAHTOA Basics
Sine, cosine, and tangent in a right triangle
Learning goal: Choose the correct trig ratio and compute \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\) from side lengths.
Key idea
In a right triangle, pick one acute angle \(\theta\). Relative to \(\theta\): opposite is across from \(\theta\), adjacent touches \(\theta\) (but is not the hypotenuse), and the hypotenuse is across from the \(90^\circ\) angle. Then: \[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}. \] This is the classic mnemonic SOHCAHTOA.
Worked example
Example: If the adjacent side is \(12\) and the hypotenuse is \(13\), what is \(\cos\theta\)?
Use \(\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}\): \[ \cos\theta=\frac{12}{13}. \]
Try it
Try it 1: In a right triangle with hypotenuse \(10\) and opposite side \(5\), what is \(\sin\theta\)?
Hint: \(\sin\theta=\dfrac{\text{opp}}{\text{hyp}}=\dfrac{5}{10}\), then simplify.
Try it 2: In a right triangle, if opposite \(=9\) and adjacent \(=12\), what is \(\tan\theta\)?
Hint: \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}=\dfrac{9}{12}\), then simplify.
Opposite/adjacent depend on which acute angle you call \(\theta\).
Reciprocal Ratios
Secant, cosecant, and cotangent
Learning goal: Switch between \(\sin,\cos,\tan\) and their reciprocals \(\csc,\sec,\cot\) using side ratios.
Key idea
The reciprocal trig ratios are: \[ \csc\theta=\frac{1}{\sin\theta}=\frac{\text{hyp}}{\text{opp}},\quad \sec\theta=\frac{1}{\cos\theta}=\frac{\text{hyp}}{\text{adj}},\quad \cot\theta=\frac{1}{\tan\theta}=\frac{\text{adj}}{\text{opp}}. \] If you can compute \(\sin\), \(\cos\), or \(\tan\), you can get \(\csc\), \(\sec\), or \(\cot\) immediately by flipping the fraction.
Worked example
Example: A right triangle has opposite \(8\), adjacent \(15\), and hypotenuse \(17\). What is \(\sec\theta\)?
Learning goal: Combine the Pythagorean theorem with trig ratios to find unknown sides and new ratios.
Key idea
Many right-triangle ratio problems are a two-step process: (1) find a missing side using the Pythagorean theorem \(a^2+b^2=c^2\), then (2) compute \(\sin\theta\), \(\cos\theta\), \(\tan\theta\) or their reciprocals. If you are given a ratio like \(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}\), you can often reconstruct the triangle.
Worked example
Example: In a right triangle, \(\cos\theta=\dfrac{8}{17}\). What is \(\sin\theta\)?
\(\cos\theta=\dfrac{\text{adj}}{\text{hyp}}=\dfrac{8}{17}\), so adjacent \(=8\) and hypotenuse \(=17\). Find the opposite side: \[ \text{opp}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15. \] Then \[ \sin\theta=\frac{\text{opp}}{\text{hyp}}=\frac{15}{17}. \]
Try it
Try it 1: If \(\cos\theta=\dfrac{5}{13}\), what is \(\sec\theta\)?
Hint: \(\sec\theta=\dfrac{1}{\cos\theta}\), so flip the fraction.
Try it 2: In a right triangle, if \(\tan\theta=2\) and the adjacent side is \(3\), what is the opposite side?
Hint: \(\tan\theta=\dfrac{\text{opp}}{\text{adj}}\). So \(\text{opp}=2\cdot 3\).
Summary
Use \(a^2+b^2=c^2\) to find missing sides when needed.
Then compute the desired trig ratio using SOHCAHTOA or reciprocals.
Special Angles
Special triangles and exact trig values
Learning goal: Recall exact trig values for \(30^\circ\), \(45^\circ\), and \(60^\circ\) using special right triangles.
Key idea
Two special right triangles give exact trig values: \(45\text{-}45\text{-}90\) has side ratio \(1:1:\sqrt{2}\), and \(30\text{-}60\text{-}90\) has side ratio \(1:\sqrt{3}:2\). From these, you can derive: \[ \sin 30^\circ=\frac12,\ \cos 30^\circ=\frac{\sqrt3}{2},\ \tan 30^\circ=\frac{\sqrt3}{3} \] \[ \sin 45^\circ=\frac{\sqrt2}{2},\ \cos 45^\circ=\frac{\sqrt2}{2},\ \tan 45^\circ=1 \] \[ \sin 60^\circ=\frac{\sqrt3}{2},\ \cos 60^\circ=\frac12,\ \tan 60^\circ=\sqrt3 \]
Worked example
Example: What is \(\tan(30^\circ)\)?
From the \(30\text{-}60\text{-}90\) triangle, \(\tan(30^\circ)=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}\).
Try it
Try it 1: What is \(\cos(45^\circ)\)?
Hint: In a \(45\text{-}45\text{-}90\) triangle, \(\cos 45^\circ=\dfrac{1}{\sqrt2}\).
Try it 2: What is \(\csc(60^\circ)\)?
Hint: \(\csc\theta=\dfrac{1}{\sin\theta}\) and \(\sin 60^\circ=\dfrac{\sqrt3}{2}\).
Summary
Special triangles give exact values at \(30^\circ\), \(45^\circ\), \(60^\circ\).
Use reciprocals to get \(\sec\), \(\csc\), \(\cot\) from \(\cos\), \(\sin\), \(\tan\).
Complementary Angles
Complementary angles and cofunction identities
Learning goal: Use the fact that acute angles in a right triangle are complementary to rewrite trig expressions.
Key idea
In any right triangle, the two acute angles add to \(90^\circ\). So if one acute angle is \(\theta\), the other is \(90^\circ-\theta\). This creates cofunction identities: \[ \sin(90^\circ-\theta)=\cos\theta,\quad \cos(90^\circ-\theta)=\sin\theta, \] \[ \tan(90^\circ-\theta)=\cot\theta,\quad \cot(90^\circ-\theta)=\tan\theta, \] \[ \sec(90^\circ-\theta)=\csc\theta,\quad \csc(90^\circ-\theta)=\sec\theta. \] A very common conversion is \(\tan(90^\circ-\theta)=\cot\theta=\dfrac{1}{\tan\theta}\).
Worked example
Example: Rewrite \(\sin(90^\circ-\theta)\) in terms of \(\theta\).
Because the angles are complementary in a right triangle: \[ \sin(90^\circ-\theta)=\cos\theta. \]
Try it
Try it 1: What is \(\tan\bigl(90^\circ-\theta\bigr)\) in terms of \(\tan\theta\)?
Hint: \(\tan(90^\circ-\theta)=\cot\theta\) and \(\cot\theta=\dfrac{1}{\tan\theta}\).
Try it 2: If \(\sin\theta=\dfrac{3}{5}\), what is \(\cot\theta\)?
Hint: If \(\sin\theta=\dfrac{3}{5}\), use a \(3\text{-}4\text{-}5\) triangle: opposite \(=3\), hypotenuse \(=5\), adjacent \(=4\). Then \(\cot\theta=\dfrac{\text{adj}}{\text{opp}}\).
Summary
Complementary angles: \(\theta\) and \(90^\circ-\theta\).
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the right-triangle ratio skill you need.
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