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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 θ\theta you care about: the hypotenuse is always opposite the right angle, the opposite side faces θ\theta, and the adjacent side touches it.

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=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=90°a^2 + b^2 = c^2 \qquad \text{and} \qquad A + B = 90°

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

These ratios only apply to a triangle that actually contains a 90°90° angle. For a non-right triangle you need the Law of Sines or the Law of Cosines instead — 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=arctan⁡ ⁣(ab)A = \arctan\!\left(\frac{a}{b}\right) or A=arcsin⁡ ⁣(ac)A = \arcsin\!\left(\frac{a}{c}\right).
  3. Get the other angle from B=90°−AB = 90° - A.

Case 2: one side and one acute angle known

  1. The other acute angle is 90°−A90° - A immediately.
  2. Pick the ratio that connects the side you have to the side you want — that is what "which trig function do I use?" really means. Known hypotenuse and want the opposite side: a=csin⁡Aa = c\sin A. Known opposite and want the hypotenuse: c=a/sin⁡Ac = a / \sin A.

Choosing the function

Write down which two sides the problem involves. Opposite and hypotenuse ⇒\Rightarrow sine. Adjacent and hypotenuse ⇒\Rightarrow cosine. Opposite and adjacent ⇒\Rightarrow tangent. Use arcsin⁡\arcsin, arccos⁡\arccos, arctan⁡\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 sin⁡30°=0.5\sin 30° = 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 sin⁡\sin where arcsin⁡\arcsin is needed: sin⁡\sin turns an angle into a ratio; arcsin⁡\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: tan⁡A=512\tan A = \frac{5}{12}, so A=arctan⁡(0.4167)≈22.62°A = \arctan(0.4167) \approx 22.62°
Step 3: Angle B=90°−22.62°=67.38°B = 90° - 22.62° = 67.38°
Step 4: Check the ratios: sin⁡A=513≈0.3846\sin A = \frac{5}{13} \approx 0.3846, cos⁡A=1213≈0.9231\cos A = \frac{12}{13} \approx 0.9231
Answer: c=13c = 13, A≈22.62°A \approx 22.62°, B≈67.38°B \approx 67.38°

Step 1: Opposite side uses sine: a=csin⁡A=20sin⁡35°a = c\sin A = 20\sin 35°
Step 2: sin⁡35°≈0.5736\sin 35° \approx 0.5736, so a≈11.47a \approx 11.47
Step 3: Adjacent side uses cosine: b=ccos⁡A=20cos⁡35°b = c\cos A = 20\cos 35°
Step 4: cos⁡35°≈0.8192\cos 35° \approx 0.8192, so b≈16.38b \approx 16.38
Step 5: Other angle: B=90°−35°=55°B = 90° - 35° = 55°
Answer: a≈11.47a \approx 11.47, b≈16.38b \approx 16.38, B=55°B = 55°

Step 1: Adjacent =8= 8, hypotenuse =17= 17, so the opposite side is 172−82=289−64=225=15\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15
Step 2: θ=arccos⁡ ⁣(817)=arccos⁡(0.4706)≈61.93°\theta = \arccos\!\left(\frac{8}{17}\right) = \arccos(0.4706) \approx 61.93°
Step 3: sin⁡θ=1517≈0.8824\sin\theta = \frac{15}{17} \approx 0.8824
Step 4: tan⁡θ=158=1.875\tan\theta = \frac{15}{8} = 1.875
Answer: θ≈61.93°\theta \approx 61.93°, sin⁡θ=1517\sin\theta = \frac{15}{17}, tan⁡θ=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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