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Right Triangle Trigonometry — Definition, Formula & Examples

Right triangle trigonometry is the study of how the angles of a right triangle relate to the lengths of its sides. If you know one acute angle and one side, you can find the other sides, and if you know two sides, you can find the angles.

For an acute angle θ\theta in a right triangle, the three primary trigonometric ratios are sin⁡θ=oppositehypotenuse\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, and tan⁡θ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}}, where the sides are named relative to θ\theta. These ratios depend only on θ\theta, not on the triangle's size, because all right triangles containing angle θ\theta are similar. The reciprocal ratios are csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, and cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}.

Key Formula

sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj\sin\theta = \frac{\text{opp}}{\text{hyp}}, \qquad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \qquad \tan\theta = \frac{\text{opp}}{\text{adj}}
Where:
  • θ\theta = An acute angle of the right triangle (not the right angle)
  • opp\text{opp} = Length of the side across from angle θ
  • adj\text{adj} = Length of the leg next to angle θ (not the hypotenuse)
  • hyp\text{hyp} = Length of the hypotenuse, the side across from the right angle

Worked Example

Problem: A right triangle has an acute angle of 30∘30^\circ and a hypotenuse of length 10. Find the lengths of the side opposite the 30∘30^\circ angle and the side adjacent to it.
Step 1: Label the sides relative to the 30∘30^\circ angle. You know the hypotenuse, and you want the opposite and adjacent sides. SOH (sine) links opposite to hypotenuse, and CAH (cosine) links adjacent to hypotenuse.
sin⁡30∘=opp10,cos⁡30∘=adj10\sin 30^\circ = \frac{\text{opp}}{10}, \qquad \cos 30^\circ = \frac{\text{adj}}{10}
Step 2: Solve for the opposite side. Multiply both sides by 10 and use sin⁡30∘=12\sin 30^\circ = \frac{1}{2}.
opp=10sin⁡30∘=10⋅12=5\text{opp} = 10\sin 30^\circ = 10 \cdot \tfrac{1}{2} = 5
Step 3: Solve for the adjacent side using cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}. Check with the Pythagorean theorem: 52+(53)2=25+75=100=1025^2 + (5\sqrt{3})^2 = 25 + 75 = 100 = 10^2.
adj=10cos⁡30∘=10⋅32=53≈8.66\text{adj} = 10\cos 30^\circ = 10 \cdot \tfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66
Answer: The opposite side is 5 and the adjacent side is 53≈8.665\sqrt{3} \approx 8.66.

Why It Matters

In physics, you use sine and cosine to split a force or velocity into horizontal and vertical components. Surveyors and engineers find heights and distances they cannot measure directly, such as a building's height from its shadow. These right triangle ratios also lead directly to the unit circle definitions of trig functions in precalculus.

Common Mistakes

Mistake: Treating 'opposite' and 'adjacent' as fixed sides of the triangle.
Correction: Those labels depend on which acute angle you are using. The side opposite one acute angle is adjacent to the other. Only the hypotenuse stays the same. Relabel the sides every time you switch angles.

Related Terms