Right Triangle Calculator

Find missing sides, angles and trig ratios in a right triangle, step by step
Solve the right triangle with legs 5 and 12
Right triangle with hypotenuse 20 and angle 35 degrees
Find the angle whose cosine is 8/17
Find the trig ratios for a 9-40-41 right triangle

The Three Trig Ratios

In a right triangle, label the sides relative to the acute angle ฮธ\theta you care about: the hypotenuse is always opposite the right angle, the opposite side faces ฮธ\theta, and the adjacent side touches it.

sinโกฮธ=oppositehypotenuse,cosโกฮธ=adjacenthypotenuse,tanโกฮธ=oppositeadjacent\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}

That is SOH-CAH-TOA. Two more facts complete the toolkit:

a2+b2=c2andA+B=90ยฐa^2 + b^2 = c^2 \qquad \text{and} \qquad A + B = 90ยฐ

for legs a,ba, b, hypotenuse cc and acute angles A,BA, B.

These ratios only apply to a triangle that actually contains a 90ยฐ90ยฐ angle. For a non-right triangle you need the Law of Sines or the Law of Cosines instead โ€” SOH-CAH-TOA will give a wrong answer there, not an error message.

How to Solve a Right Triangle

You need two pieces of information besides the right angle, and at least one of them must be a side. Match your case:

Case 1: two sides known

  1. Find the third side with a2+b2=c2a^2 + b^2 = c^2.
  2. Find one acute angle with an inverse ratio, e.g. A=arctanโกโ€‰โฃ(ab)A = \arctan\!\left(\frac{a}{b}\right) or A=arcsinโกโ€‰โฃ(ac)A = \arcsin\!\left(\frac{a}{c}\right).
  3. Get the other angle from B=90ยฐโˆ’AB = 90ยฐ - A.

Case 2: one side and one acute angle known

  1. The other acute angle is 90ยฐโˆ’A90ยฐ - A immediately.
  2. Pick the ratio that connects the side you have to the side you want โ€” that is what "which trig function do I use?" really means. Known hypotenuse and want the opposite side: a=csinโกAa = c\sin A. Known opposite and want the hypotenuse: c=a/sinโกAc = a / \sin A.

Choosing the function

Write down which two sides the problem involves. Opposite and hypotenuse โ‡’\Rightarrow sine. Adjacent and hypotenuse โ‡’\Rightarrow cosine. Opposite and adjacent โ‡’\Rightarrow tangent. Use arcsinโก\arcsin, arccosโก\arccos, arctanโก\arctan when the unknown is the angle.

Common Mistakes to Avoid

  • Calculator in the wrong mode: degree answers from radian mode are the single most common source of wrong results. Check that sinโก30ยฐ=0.5\sin 30ยฐ = 0.5 before you start.
  • Fixed labelling: "opposite" and "adjacent" swap when you switch which acute angle you are working from. Only the hypotenuse never moves.
  • Using sinโก\sin where arcsinโก\arcsin is needed: sinโก\sin turns an angle into a ratio; arcsinโก\arcsin turns a ratio back into an angle.
  • Applying a2+b2=c2a^2 + b^2 = c^2 with cc as a leg: cc must be the hypotenuse, the longest side.
  • Rounding too early: carry full precision through intermediate steps and round only the final answer, or angles drift by a tenth of a degree.

Examples

Step 1: Hypotenuse: c=52+122=25+144=169=13c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
Step 2: Angle AA: tanโกA=512\tan A = \frac{5}{12}, so A=arctanโก(0.4167)โ‰ˆ22.62ยฐA = \arctan(0.4167) \approx 22.62ยฐ
Step 3: Angle B=90ยฐโˆ’22.62ยฐ=67.38ยฐB = 90ยฐ - 22.62ยฐ = 67.38ยฐ
Step 4: Check the ratios: sinโกA=513โ‰ˆ0.3846\sin A = \frac{5}{13} \approx 0.3846, cosโกA=1213โ‰ˆ0.9231\cos A = \frac{12}{13} \approx 0.9231
Answer: c=13c = 13, Aโ‰ˆ22.62ยฐA \approx 22.62ยฐ, Bโ‰ˆ67.38ยฐB \approx 67.38ยฐ

Step 1: Opposite side uses sine: a=csinโกA=20sinโก35ยฐa = c\sin A = 20\sin 35ยฐ
Step 2: sinโก35ยฐโ‰ˆ0.5736\sin 35ยฐ \approx 0.5736, so aโ‰ˆ11.47a \approx 11.47
Step 3: Adjacent side uses cosine: b=ccosโกA=20cosโก35ยฐb = c\cos A = 20\cos 35ยฐ
Step 4: cosโก35ยฐโ‰ˆ0.8192\cos 35ยฐ \approx 0.8192, so bโ‰ˆ16.38b \approx 16.38
Step 5: Other angle: B=90ยฐโˆ’35ยฐ=55ยฐB = 90ยฐ - 35ยฐ = 55ยฐ
Answer: aโ‰ˆ11.47a \approx 11.47, bโ‰ˆ16.38b \approx 16.38, B=55ยฐB = 55ยฐ

Step 1: Adjacent =8= 8, hypotenuse =17= 17, so the opposite side is 172โˆ’82=289โˆ’64=225=15\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15
Step 2: ฮธ=arccosโกโ€‰โฃ(817)=arccosโก(0.4706)โ‰ˆ61.93ยฐ\theta = \arccos\!\left(\frac{8}{17}\right) = \arccos(0.4706) \approx 61.93ยฐ
Step 3: sinโกฮธ=1517โ‰ˆ0.8824\sin\theta = \frac{15}{17} \approx 0.8824
Step 4: tanโกฮธ=158=1.875\tan\theta = \frac{15}{8} = 1.875
Answer: ฮธโ‰ˆ61.93ยฐ\theta \approx 61.93ยฐ, sinโกฮธ=1517\sin\theta = \frac{15}{17}, tanโกฮธ=158\tan\theta = \frac{15}{8}

Frequently Asked Questions

Identify the two sides the problem mentions relative to your angle. Opposite with hypotenuse means sine, adjacent with hypotenuse means cosine, opposite with adjacent means tangent. If the unknown is the angle rather than a side, use the inverse version of that same function.

Two values in addition to the right angle, and at least one must be a side length. Two angles alone fix the shape but not the size, so every side would remain unknown.

No. SOH-CAH-TOA is defined by the right angle. For other triangles use the Law of Sines a/sin A = b/sin B, or the Law of Cosines c^2 = a^2 + b^2 - 2ab cos C, which reduces to the Pythagorean theorem when C = 90 degrees.

Almost always because it is in radian or gradian mode when the question is in degrees. Test it with sin(30): you should get 0.5 in degree mode. Also check that any ratio you feed into arcsin or arccos is between -1 and 1, since values outside that range have no solution.

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