How To Find Angle Of Triangle Given 2 Sides

You're staring at a triangle. Consider this: you know the length of two sides. You need the angle between them. Consider this: maybe it's a roof truss, a piece of code calculating a projectile path, or just a homework problem due tomorrow. Or maybe the angle opposite one of them.

Here's the thing nobody tells you upfront: two sides alone are never enough.

Not unless you sneak in a third piece of information. A right angle. The included angle. The third side. Something. Without that extra constraint, the triangle isn't a triangle yet — it's a hinge. Worth adding: it swings. The angle you're hunting could be 10 degrees or 170 degrees, and both would fit the two lengths you're holding.

So the real question isn't "how do I find the angle?" It's "what else* do I know?"

What Is the Problem, Really

A triangle has six parts: three sides, three angles. That's why standard notation labels sides a, b, c and the opposite angles A, B, C. Side a sits across from angle A, and so on.

If you only have two sides — say a and b — you have an infinite family of possible triangles. And imagine two sticks connected at one end by a pivot. So naturally, open the pivot, close it. Every position is a valid triangle with those two side lengths. Day to day, the third side stretches and shrinks. The angles shift.

To lock it down, you need one more fact. That fact usually falls into one of three buckets:

1. It's a right triangle. (One angle is 90°.)
2. You know the angle between the two sides.* (SAS — Side-Angle-Side.)
3. You know an angle opposite one of the known sides.* (SSA — Side-Side-Angle, the notorious "ambiguous case.")
4. You know the third side. (SSS — Side-Side-Side, but then you'd have three sides, not two.)

This article walks through the first three. They cover 99% of the real-world scenarios where someone says "I have two sides, find the angle."

Why It Matters

You might be a carpenter cutting rafters. A surveyor staking a property line. Consider this: a game developer calculating line-of-sight. A student staring at a diagram where only two lengths are labeled.

In every case, guessing the missing constraint leads to disaster. Because of that, cut the wrong angle on a $400 piece of timber and you're buying another. Code the wrong quadrant check in a physics engine and your character clips through the floor.

The math isn't hard. The discipline* is hard — slowing down long enough to identify which case you're actually in before reaching for a formula.

How It Works: Three Scenarios, Three Toolkits

Scenario 1: The Right Triangle (Your Best Friend)

If you see a little square in the corner, or the problem says "right triangle," stop overthinking. You have SOH CAH TOA.

Label the sides relative to the angle you want (call it θ):

• Opposite = the side across from θ
• Adjacent = the side next to θ that isn't the hypotenuse
• Hypotenuse = the longest side, always opposite the right angle

Now pick the ratio that uses the two sides you have*:

• Sine = Opposite / Hypotenuse
• Cosine = Adjacent / Hypotenuse
• Tangent = Opposite / Adjacent

Example

You have a ladder (hypotenuse = 10 ft) leaning against a wall. The base sits 6 ft from the wall. You want the angle the ladder makes with the ground.

You know Adjacent (6) and Hypotenuse (10). That's Cosine.

cos(θ) = 6/10 = 0.6

θ = cos⁻¹(0.6) ≈ 53.13°

The Trap

Your calculator must* be in degree mode. Not radians. Not gradians. Degrees. I've seen smart people waste an hour debugging perfect code because Math.acos(0.6) returned 0.927 (radians) and they treated it like degrees.

Also: inverse trig functions (sin⁻¹, cos⁻¹, tan⁻¹) only return principal values.

• sin⁻¹ and tan⁻¹ give angles between -90° and 90°.
• cos⁻¹ gives angles between 0° and 180°.

In a right triangle, your unknown angle is always acute (0° to 90°), so you're safe. But the moment you leave right-triangle land, this becomes a minefield.

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Scenario 2: SAS — You Know the Included Angle

Scenario 3: SSA — The Ambiguous Case (The Trickiest Puzzle)

When you know two sides and a non-included angle, the situation becomes unpredictable. This is the Side-Side-Angle (SSA) case, often called the "ambiguous case" because it can yield zero, one, or two valid triangles. The key is to determine how the given sides and angle interact.

The SSA Workflow

1. Label the triangle: Call the known sides a and b, and the known angle A (opposite side a).
2. Use the Law of Sines to find the angle opposite the unknown side:
   $ \frac{\sin A}{a} = \frac{\sin B}{b} \implies \sin B = \frac{b \cdot \sin A}{a} $
   • If $\frac{b

Scenario 3: SSA — The Ambiguous Case (The Trickiest Puzzle)

When you know two sides and a non-included angle, the situation becomes unpredictable. This is the Side-Side-Angle (SSA) case, often called the "ambiguous case" because it can yield zero, one, or two valid triangles. The key is to determine how the given sides and angle interact.

The SSA Workflow

1. Label the triangle: Call the known sides a and b, and the known angle A (opposite side a).
2. **Use the Law

… of Sines to find the angle opposite the unknown side:

[ \frac{\sin A}{a}= \frac{\sin B}{b};\Longrightarrow; \sin B = \frac{b\sin A}{a}. ]

Now evaluate the value of (\sin B):

Condition on (\displaystyle \frac{b\sin A}{a})	Interpretation	Number of possible triangles
> 1	The sine of an angle cannot exceed 1 → no real angle (B) satisfies the equation.	0 (no triangle can be formed)
= 1	(\sin B = 1) → (B = 90^\circ). That said,	1 (unique right triangle)
< 1	Two supplementary angles have the same sine: (B_1 = \sin^{-1}! Day to day, the known angle (A) and side lengths force a right triangle. That's why \left(\frac{b\sin A}{a}\right)) and (B_2 = 180^\circ - B_1). Both are mathematically possible, but only those that keep the triangle’s interior angles less than (180^\circ) survive.

To decide between one and two solutions in the “< 1” case, check the sum (A + B_i):

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• If (A + B_1 \ge 180^\circ), then (B_1) is inadmissible (the remaining angle would be ≤ 0).
• If (A + B_2 \ge 180^\circ), then (B_2) is inadmissible.

When both sums are < 180°, each yields a valid triangle, giving the classic ambiguous case with two distinct solutions.

SSA Example

Suppose (a = 8) cm, (b = 10) cm, and the known angle (A = 30^\circ) (opposite side (a)).

[ \sin B = \frac{b\sin A}{a}= \frac{10\cdot\sin30^\circ}{8}= \frac{10\cdot0.5}{8}=0.625. ]

Since (0.625<1), we compute

[ B_1 = \sin^{-1}(0.This leads to 625) \approx 38. 68^\circ,\qquad B_2 = 180^\circ-38.Practically speaking, 68^\circ = 141. 32^\circ.

Check the sums with (A):

• (A + B_1 = 30^\circ + 38.68^\circ = 68.68^\circ < 180^\circ) → valid.
• (A + B_2 = 30^\circ + 141.32^\circ = 171.32^\circ < 180^\circ) → also valid (the leftover angle (C = 180^\circ - A - B_2 \approx 8.68^\circ) is positive).

Thus two triangles satisfy the given SSA data: one with angles ((30^\circ, 38.68^\circ, 111.Even so, 32^\circ)) and another with ((30^\circ, 141. 32^\circ, 8.68^\circ)).

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Scenario 2: SAS — You Know the Included Angle

When two sides and the angle between them are known (Side‑Angle‑Side), the Law of Cosines is the direct route to the third side, after which the Law of Sines (or again Cosines) yields the remaining angles.

1. Label the known sides as (s_1) and (s_2) and the included angle as (\gamma).

2. Compute the unknown side (s_3) via

   [ s_3^{2}= s_1^{

3. Compute the unknown side (s_3) via

[ s_3^{2}= s_1^{2}+s_2^{2}-2s_1s_2\cos\gamma . ]

Take the positive square root (since side lengths are positive) to obtain (s_3).

3. Find the remaining angles – you can use either the Law of Sines or the Law of Cosines again. A common approach is to apply the Law of Sines to the angle opposite one of the known sides, for instance

[ \frac{\sin\alpha}{s_1}= \frac{\sin\gamma}{s_3}\quad\Longrightarrow\quad \alpha = \sin^{-1}!\left(\frac{s_1\sin\gamma}{s_3}\right), ]

and then obtain the third angle by subtracting the two known angles from (180^\circ): (\beta = 180^\circ - \alpha - \gamma). (If you instead use the Law of Cosines for the angles, you avoid any ambiguity from the inverse sine.)

Because the two sides and the included angle uniquely determine the triangle, the SAS case always yields exactly one solution.

SAS Example

Let (s_1 = 5) cm, (s_2 =

Let (s_1 = 5) cm, (s_2 = 7) cm, and the included angle (\gamma = 60^\circ).

First, compute the unknown side (s_3) using the Law of Cosines:

[ s_3^2 = s_1^2 + s_2^2 - 2s_1s_2\cos\gamma = 5^2 + 7^2 - 2(5)(7)\cos(60^\circ) ]

[ s_3^2 = 25 + 49 - 70(0.5) = 74 - 35 = 39 ]

[ s_3 = \sqrt{39} \approx 6.245 \text{ cm} ]

Next, find one of the remaining angles using the Law of Sines:

[ \frac{\sin\alpha}{s_1} = \frac{\sin\gamma}{s_3} \Rightarrow \sin\alpha = \frac{s_1 \sin\gamma}{s_3} = \frac{5 \cdot \sin(60^\circ)}{6.245} ]

[ \sin\alpha = \frac{5 \cdot 0.245} \approx \frac{4.866}{6.Still, 33}{6. 245} \approx 0.

[ \alpha = \sin^{-1}(0.693) \approx 43.85^\circ ]

Finally, find the third angle:

[ \beta = 180^\circ - \alpha - \gamma = 180^\circ - 43.85^\circ - 60^\circ = 76.15^\circ ]

The triangle has sides approximately (5) cm, (7) cm, and (6.On the flip side, 245) cm, with angles (60^\circ), (43. Here's the thing — 85^\circ), and (76. 15^\circ).

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Scenario 3: SSS — All Three Sides Known

When all three sides of a triangle are known (Side-Side-Side), the Law of Cosines is used directly to compute each angle. This method avoids any ambiguity since there is no inverse sine involved.

For a triangle with sides (a), (b), and (c), the angle opposite side (a) (denoted as (A)) can be found using:

[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} ]

Similar formulas apply for angles (B) and (C):

[ \cos B = \frac{a^2 + c^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} ]

Each angle is then obtained by taking the inverse cosine.

SSS Example

Let (a = 9) cm, (b = 12) cm, and (c = 15) cm.

Compute angle (A):

[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{12^2 + 15^2 - 9^2}{2(12)(15)} = \frac{144 + 225 - 81}{360} = \frac{288}{360} = 0.8 ]

[ A = \cos^{-1}(0.8) \approx 36.87^\circ ]

Compute angle (B):

[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{9^2 + 15^2 - 12^2}{2(9)(15)} = \frac{81 + 225 - 144}{270} = \frac{162}{270} = 0.6 ]

[ B = \cos^{-1}(0.6) \approx 53.13^\circ ]

Finally, compute angle (C):

[ C = 180^\circ - A - B = 180^\circ - 36.87^\circ - 53.13^\circ = 90^\circ ]

This confirms that the triangle is right-angled at (C), which makes sense given the side lengths (9), (12), and (15) form a scaled version of the Pythagorean triple ((3, 4, 5)).

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Conclusion

Solving triangles based on different given conditions requires a clear understanding of which trigonometric laws to apply and when each is most appropriate. The SSA case introduces potential ambiguity that must be carefully checked, while SAS and SSS cases yield unique solutions through the Law of Cosines. Whether dealing with real-world applications like surveying or engineering design, mastering these techniques ensures accurate determination of unknown sides and angles in any triangle configuration.