How to Solve a Right Triangle

A right triangle has one 90° angle, and that single fixed corner is what makes it so predictable to work with. To solve a triangle means to pin down every measurement in it: all three side lengths and all three angles. The good news is that in a right triangle you rarely need to start with much. Two extra facts on top of the right angle are almost always enough to recover everything else.

The three sides, named from an angle

Trigonometry does not give the sides fixed names like "left" or "bottom." Instead it labels them relative to whichever acute angle you are focused on. The hypotenuse is the longest side, sitting directly across from the right angle; it never changes no matter which angle you pick. The opposite side is the one facing your chosen angle across the triangle, and the adjacent side is the one that forms the angle together with the hypotenuse. Shift your attention to the other acute angle and the opposite and adjacent sides trade roles, while the hypotenuse stays exactly where it is.

Finding a missing side

Three ratios connect an angle to a pair of its sides:

• sin θ = opposite / hypotenuse
• cos θ = adjacent / hypotenuse
• tan θ = opposite / adjacent

The trick is to choose the ratio that mentions the side you already know together with the side you want. If you know the hypotenuse and want the opposite side, sine is your tool. If you know the adjacent side and want the opposite one, tangent connects them. Once the right ratio is chosen, a single multiplication or division isolates the unknown length.

Finding a missing angle

When you know two sides but no acute angle, run the ratios in reverse using the inverse functions sin⁻¹, cos⁻¹, and tan⁻¹. For instance, if the opposite side is 3 and the adjacent side is 4, then tan θ = 3/4 = 0.75, so θ = tan⁻¹(0.75) ≈ 36.87°. The inverse function simply asks the question "which angle has this ratio?"

Two rules that tie it together

The Pythagorean theorem, a² + b² = c² with c the hypotenuse, lets you find any side from the other two without touching an angle at all. The angle-sum rule says the three interior angles of any triangle total 180°. Because one of them is already 90°, the two acute angles must add up to 90° on their own, which means each is the complement of the other. Together these two facts often let you finish a triangle with pure arithmetic once trig has given you a foothold.

Worked example

Suppose one acute angle measures 35° and the hypotenuse is 12 units long. Find the remaining two sides and the third angle.

1. Third angle. 180° − 90° − 35° = 55°.
2. Opposite side. Since sin 35° ≈ 0.5736, the opposite side is 12 × 0.5736 ≈ 6.88.
3. Adjacent side. Since cos 35° ≈ 0.8192, the adjacent side is 12 × 0.8192 ≈ 9.83.

It is always worth checking the result against Pythagoras: 6.88² + 9.83² ≈ 47.3 + 96.6 = 143.9, which rounds neatly to 12² = 144. The small gap is nothing more than rounding along the way, so the triangle is fully solved: sides 6.88, 9.83, and 12, with angles 35°, 55°, and 90°.

Two starting situations

In practice a right-triangle problem lands in one of two shapes. Either you are given an angle plus a side, in which case a trig ratio finds a second side and the angle-sum rule finds the last angle, as in the example above. Or you are given two sides and no acute angle, in which case Pythagoras finds the third side while an inverse trig function recovers an angle. Learning to recognise which situation you are in is half the battle, because it tells you immediately whether to reach for a ratio or for the theorem first.

Why a lone angle is never enough

It is worth pausing on why the right angle plus one more angle can never finish a triangle, no matter how precisely you know it. Picture two right triangles that share the same 35° corner — one with a 12-unit hypotenuse, another with a hypotenuse a thousand times longer. Both have identical angles (35°, 55°, 90°), because angles alone describe a shape, not a size. Every ratio of sides is fixed by the angles, but the actual lengths are not. That is exactly why solving a right triangle always needs at least one length: without one, you have correctly identified the shape of infinitely many similarly-shaped triangles, of every possible size, and no way to pick out the one you actually have.

A second worked example: two sides, no angle

Now try the other common starting point — both legs are known and neither acute angle is given. Suppose the two legs measure 9 and 12 units.

1. Hypotenuse, by Pythagoras. c² = 9² + 12² = 81 + 144 = 225, so c = √225 = 15.
2. Angle A, opposite the leg of 9. tan A = 9/12 = 0.75, so A = tan⁻¹(0.75) ≈ 36.87°.
3. Angle B, the complement. B = 90° − 36.87° = 53.13°.

That gives the fully solved triangle: sides 9, 12, and 15, with angles 36.87°, 53.13°, and 90°. Notice this is nothing more than the familiar 3-4-5 right triangle scaled up by a factor of 3 — 9 is 3 × 3, 12 is 3 × 4, and 15 is 3 × 5 — which is exactly the similar-triangles point from above: scaling every side by the same factor leaves every angle untouched.

Area and perimeter come along for free

Once all three sides and both acute angles are known, two more measurements fall out with no extra trig at all. The perimeter is simply the sum of the three sides: for the 9-12-15 triangle, 9 + 12 + 15 = 36. The area uses the fact that the two legs are already perpendicular to each other, so they can stand in directly for "base" and "height": area = ½ × a × b = ½ × 9 × 12 = 54 square units. This shortcut only works because the right angle sits between the two legs — in a general triangle without a right angle, finding the area takes an extra step (typically ½ab sin C, using the angle between the two sides), which is exactly the kind of case the Law of Sines and Law of Cosines are built to handle.

This is also why right triangles turn up so often outside the classroom: a ladder leaning on a wall, a ramp rising to a fixed height, or a roof pitch expressed as an angle all reduce to exactly the shape above — a base, a height, and a slanted third side connecting them, with the ground and the wall meeting at 90°. Whatever the setting, the same two facts (one side plus one more piece of information, angle or side) are always enough.

A third shape: a leg and the hypotenuse

The two examples above cover "angle plus hypotenuse" and "two legs with no angle." A third common shape is a single leg together with the hypotenuse and no angle at all. Suppose leg b = 12 and hypotenuse c = 20. The other leg comes from Pythagoras: a² = c² − b² = 400 − 144 = 256, so a = 16 — another whole-number triple, this time 12-16-20 (a scaled-up 3-4-5). Since b sits opposite B, it is adjacent to the other acute angle, A, which makes cosine (CAH) the direct route to A: A = cos⁻¹(12/20) = cos⁻¹(0.6) ≈ 53.13°, and B = 90° − 53.13° = 36.87°. The calculator confirms it exactly: A ≈ 53.13°, B ≈ 36.87°, area 96, and perimeter 48. A guard rail worth knowing about here mirrors the one from two legs: the given leg must be strictly shorter than the hypotenuse, since a leg can never reach or exceed the hypotenuse in a right triangle — feeding in b = 22 against c = 20 has no solution, and asking for one should fail loudly rather than return a nonsense angle.

The four legitimate starting points

Every solvable right-triangle problem reduces to one of four combinations of "two knowns plus the right angle": an acute angle with the hypotenuse, an acute angle with a leg, two legs, or a leg with the hypotenuse. Each one reaches the same destination by a slightly different route — sine or cosine first for an angle-and-hypotenuse pair, tangent first for an angle-and-leg pair, Pythagoras first for two sides. The two examples worked through above cover the "angle and hypotenuse" and "two legs" cases; Solving a Right Triangle From Any Two Given Values walks through all four side by side with fresh numbers, so you can see exactly which ratio or theorem leads off in each situation.

Two mistakes worth guarding against

Two slips account for most wrong answers here. The first is a calculator mode error: if your calculator is set to radians while you are entering degrees (or the reverse), every trig value comes out wrong, and the triangle you compute will violate the Pythagorean theorem by a wide margin rather than by a rounding whisker. If a check like the one above is off by more than a percent or so, suspect the mode before you suspect your method. The second is swapping the two acute angles' opposite sides — mixing up which leg is "opposite A" and which is "opposite B" flips the whole triangle into its mirror image. Labelling the triangle before you write a single ratio, exactly as recommended for SOHCAHTOA, heads this off completely.

One practical caution: before trusting any number, make sure your calculator is set to degrees and not radians. A triangle that comes out wildly wrong is very often a mode error rather than a mistake in your method.

Once you have the pattern, the whole process becomes mechanical. The right triangle calculator runs both situations above instantly and will not let you forget a required input, and SOHCAHTOA Explained digs deeper into choosing the right ratio. When the right angle disappears entirely, the method changes — see Law of Sines vs Law of Cosines for what replaces it. Try it yourself with the calculators on SinCosTan and watch the numbers fall into place.