What is the ambiguous case (SSA)?

When given two sides and a non-included angle, the setup may yield two possible angles, resulting in two valid triangles. If the calculated sine value exceeds 1, no triangle exists.

How do I know if the second SSA triangle is valid?

Calculate the supplementary angle ($ heta_2 = 180^\circ - heta_1$). If the sum of $ heta_2$ and the originally given angle is less than $180^\circ$, the second triangle is valid.

Can I use the law of sines for right triangles?

Yes, but standard right-triangle trigonometry (SOH CAH TOA) is more efficient, as $\sin(90^\circ) = 1$ simplifies the equations immediately.

How to solve a triangle with the law of sines

The law of sines states that the ratio of a triangle's side length to the sine of its opposite angle is constant for all three sides.

Apply this method to solve oblique triangles when you are given two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA).

The setup

The standard equation is $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Sketch the triangle, labeling vertices with uppercase letters ( $A, B, C$ ) and the opposite sides with corresponding lowercase letters ( $a, b, c$ ). Identify the given variable pair (a side and its opposite angle) to establish the known ratio.

The steps

Select two terms from the law of sines equation containing the known ratio and exactly one unknown variable.
Isolate the unknown variable algebraically.
Calculate the value. If solving for an angle, use the inverse sine function ( $\arcsin$ ).
Use the triangle angle sum theorem ( $A + B + C = 180^\circ$ ) to find the third angle, if needed.
Repeat steps 1-3 to find the final missing side.

Checking the result

Verify that the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. Ensure the sum of any two sides is strictly greater than the third side.

Common errors

The most frequent error occurs in the SSA configuration, known as the ambiguous case. The inverse sine function only returns angles in the first quadrant ( $\leq 90^\circ$ ). You must manually check if the obtuse supplement ( $180^\circ - heta$ ) also forms a valid triangle.

Worked example

Solve the triangle given $A = 30^\circ$ , $B = 45^\circ$ , and side $a = 10$ . Find $C$ , $b$ , and $c$ .

$C = 180^\circ - 30^\circ - 45^\circ = 105^\circ$

$\frac{a}{\sin A} = \frac{b}{\sin B}$ $b = \frac{a \sin B}{\sin A}$ $b = \frac{10 \sin 45^\circ}{\sin 30^\circ} = \frac{10(\frac{\sqrt{2}}{2})}{0.5} = 10\sqrt{2} \approx 14.14$

$\frac{a}{\sin A} = \frac{c}{\sin C}$ $c = \frac{a \sin C}{\sin A}$ $c = \frac{10 \sin 105^\circ}{\sin 30^\circ} = \frac{10(\frac{\sqrt{6}+\sqrt{2}}{4})}{0.5} = 5(\sqrt{6}+\sqrt{2}) \approx 19.32$

FAQ

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References: OpenStax Algebra and Trigonometry, Chapter 10 · Stewart Precalculus, Chapter 6