Laws of Sines & Cosines

Master the two most powerful tools for solving any triangle: the Law of Sines and the Law of Cosines.

Precalculus

What is an Oblique Triangle?

An oblique triangle is any triangle that does not contain a right angle. Since SOHCAHTOA only works for right triangles, we need different tools: the Law of Sines and the Law of Cosines.

The Law of Sines

In any triangle with sides \(a, b, c\) opposite angles \(A, B, C\):

The ratio of each side length to the sine of its opposite angle is constant throughout the triangle.

The Law of Cosines

For any triangle with sides \(a, b, c\) opposite angles \(A, B, C\):

When to Use Each Law

Law of Sines

AAS: Two angles + non-included side
ASA: Two angles + included side
SSA: Two sides + non-included angle (ambiguous case)

Law of Cosines

SAS: Two sides + included angle
SSS: All three sides known
(No ambiguity!)

The Ambiguous Case (SSA)

When given two sides and a non-included angle (SSA), the Law of Sines can yield 0, 1, or 2 triangles. This happens because \(\sin\) has two solutions in \((0^\circ, 180^\circ)\).

No Triangle

If \(\sin B > 1\), no triangle exists.

One Triangle

If \(\sin B = 1\) or the supplement yields an impossible angle sum.

Two Triangles

If both \(B\) and \(180^\circ - B\) produce valid triangles.

Step-by-Step Approach

1

Identify what you know. Label sides \(a, b, c\) and angles \(A, B, C\). Each side is opposite its corresponding angle.

2

Choose the right law. SAS or SSS: use Law of Cosines. AAS, ASA, or SSA: use Law of Sines.

3

Solve for the unknown. Plug in known values and solve algebraically. For cosines, isolate the squared term first.

4

Find remaining parts. Use \(A + B + C = 180^\circ\) and either law to find the rest. Check for ambiguous cases if SSA.

5

Verify. Check that all angles sum to \(180^\circ\) and that the longest side is opposite the largest angle.

Example 1: Law of Sines (AAS)

In \(\triangle ABC\): \(A = 40^\circ\), \(B = 75^\circ\), and \(a = 12\). Find side \(b\).

Example 2: Law of Cosines (SAS)

In \(\triangle ABC\): \(a = 8\), \(b = 11\), and \(C = 60^\circ\). Find side \(c\).

Example 3: SSS to Find Angles

In \(\triangle ABC\): \(a = 7\), \(b = 10\), \(c = 13\). Find angle \(C\).

Area with Sine

When you know two sides and the included angle, you can find the area without needing the height:

Proving the Law of Sines

1

Drop an altitude \(h\) from vertex \(C\) to side \(c\), creating two right triangles.

2

In the right triangle on the left: \(\sin A = \dfrac{h}{b}\), so \(h = b\sin A\).

3

In the right triangle on the right: \(\sin B = \dfrac{h}{a}\), so \(h = a\sin B\).

4

Set them equal: \(b\sin A = a\sin B\), so \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\).

5

Repeat by dropping an altitude from \(A\) to side \(a\) to include \(\dfrac{c}{\sin C}\).

Proving the Law of Cosines

1

Place vertex \(C\) at the origin with side \(a\) along the positive \(x\)-axis. Then \(B = (a, 0)\) and \(A = (b\cos C,\; b\sin C)\).

2

Apply the distance formula for side \(c\) (from \(A\) to \(B\)):

3

Expand: \(c^2 = b^2\cos^2 C - 2ab\cos C + a^2 + b^2\sin^2 C\)

4

Simplify using \(\cos^2 C + \sin^2 C = 1\): \(c^2 = a^2 + b^2 - 2ab\cos C\). ✓

Common Mistakes to Avoid

✕ Wrong Pairing

In Law of Sines, mismatching a side with an angle that is not opposite to it (e.g., using \(\frac{a}{\sin B}\)).

✕ Forgetting the Ambiguous Case

With SSA, not checking whether \(180^\circ - B\) also creates a valid triangle.

✕ Calculator in Wrong Mode

Using radians instead of degrees (or vice versa). Always verify your calculator's angle mode.

Sines vs. Cosines: Summary

	Law of Sines	Law of Cosines
Formula	\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)	\(c^2 = a^2 + b^2 - 2ab\cos C\)
Use When	AAS, ASA, SSA	SAS, SSS
Ambiguity?	Yes (SSA)	No
Finds	Sides or angles via proportions	Missing side, or angles via inverse cosine
Special Case	Relates to the circumscribed circle	Reduces to Pythagorean Theorem

Question 1 of 5

In \(\triangle ABC\), \(A = 30^\circ\), \(B = 45^\circ\), and \(a = 10\). Using the Law of Sines, what is \(b\)?

\(\dfrac{10}{\sin 30^\circ} = \dfrac{b}{\sin 45^\circ}\). So \(b = \dfrac{10 \cdot \sin 45^\circ}{\sin 30^\circ} = \dfrac{10 \times 0.7071}{0.5} = 14.14\).

Question 2 of 5

Two sides of a triangle are \(a = 5\) and \(b = 7\) with included angle \(C = 120^\circ\). What is \(c\)?

\(c^2 = 25 + 49 - 2(5)(7)\cos 120^\circ = 74 - 70(-0.5) = 74 + 35 = 109\). So \(c = \sqrt{109} \approx 10.44\).

Question 3 of 5

A triangle has sides \(a = 6\), \(b = 8\), \(c = 10\). What is angle \(C\)?

\(\cos C = \dfrac{36 + 64 - 100}{2(6)(8)} = \dfrac{0}{96} = 0\). So \(C = \cos^{-1}(0) = 90^\circ\). This is a 6-8-10 right triangle (a scaled 3-4-5)!

Question 4 of 5

What is the area of a triangle with sides \(a = 9\), \(b = 12\), and included angle \(C = 30^\circ\)?

\(\text{Area} = \dfrac{1}{2}(9)(12)\sin 30^\circ = \dfrac{1}{2}(108)(0.5) = 27\). Remember: \(\sin 30^\circ = 0.5\).

Question 5 of 5

Given \(A = 40^\circ\), \(a = 10\), and \(b = 14\) (SSA), when you solve \(\sin B = \dfrac{14\sin 40^\circ}{10}\), you get \(\sin B \approx 0.8999\). How many triangles are possible?

Since \(\sin B \approx 0.9 < 1\), angle \(B\) exists. \(B \approx 64.1^\circ\), so \(B' = 180^\circ - 64.1^\circ = 115.9^\circ\). Check: \(A + B' = 40^\circ + 115.9^\circ = 155.9^\circ < 180^\circ\), so both are valid. Two triangles exist: this is the ambiguous case.