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Laws of Sines
& Cosines

Derive the triangle area formula \( A = \tfrac{1}{2}ab\sin C \) and master the Law of Sines and Law of Cosines to solve any triangle.

Geometry

Beyond Right Triangles

Most real-world triangles are not right triangles. Think of the triangle formed by three cities on a map, a leaning roof truss, or a satellite signal path. Standard tools like SOHCAHTOA only work when you have a \(90^\circ\) angle.

Right Triangle
Has one \(90^\circ\) angle: use basic trig ratios directly.
Oblique Triangle
No \(90^\circ\) angle: requires the Laws of Sines and Cosines.

The Auxiliary Height Trick

Start with a generic triangle \(\triangle ABC\). Drop a perpendicular from vertex \(B\) to side \(\overline{AC}\), creating height \(h\). This single construction line unlocks everything.

A
B
C
\(c\)
\(a\)
\(b\)
\(h\)
\(C\)
\(A\)

Deriving \( A = \tfrac{1}{2}ab\sin C \)

1
Standard area formula: \( A = \tfrac{1}{2} \cdot \text{base} \cdot \text{height} = \tfrac{1}{2} \cdot b \cdot h \)
2
Express \(h\) with trig: From right triangle with angle \(C\): \( \sin C = \tfrac{h}{a} \), so \( h = a\sin C \)
3
Substitute: \( A = \tfrac{1}{2} \cdot b \cdot a\sin C = \tfrac{1}{2}ab\sin C \)

The Law of Sines

Since the area of the same triangle can be computed three ways, setting any two equal and simplifying yields a powerful proportion:

\(A\)
\(B\)
\(C\)
\(a\)
\(c\)
\(b\)

Each side is proportional to the sine of its opposite angle. Use the Law of Sines when you know an angle-side pair plus one more piece.

The Law of Cosines

This generalizes the Pythagorean Theorem. When \(C = 90^\circ\), \(\cos 90^\circ = 0\), so the formula reduces to \(c^2 = a^2 + b^2\). For any other angle, the \(-2ab\cos C\) term adjusts for the triangle's shape.

SAS
Know two sides and the included angle: find the third side directly.
SSS
Know all three sides: rearrange to find any angle.

Proving the Law of Cosines

A
B
C
D
\(h\)
\(b - x\)
\(x\)
\(c\)
\(a\)
\(b\)
1
Drop altitude \(h\) from \(B\) to \(\overline{AC}\), hitting at \(D\). Let \(CD = x\), so \(AD = b - x\).
2
In \(\triangle BDC\): \( h^2 = a^2 - x^2 \) and \( x = a\cos C \).
3
In \(\triangle BDA\): \( c^2 = h^2 + (b-x)^2 = (a^2-x^2) + b^2 - 2bx + x^2 \).
4
Simplify & substitute \(x\): \( c^2 = a^2 + b^2 - 2b(a\cos C) = a^2 + b^2 - 2ab\cos C \). ✓

When to Use Which Law?

Given Information Law of Sines Law of Cosines
AAS / ASA ✓ Use this Possible but slower
SSA (ambiguous) ✓ Use this (watch for 2 solutions) Also works
SAS Cannot start here ✓ Use this
SSS Cannot start here ✓ Use this

Step-by-Step Solving Strategy

1
Sketch and label. Draw the triangle and mark every known side and angle.
2
Classify the given info: Is it AAS/ASA, SSA, SAS, or SSS?
3
Choose your law. AAS/ASA/SSA → Law of Sines. SAS/SSS → Law of Cosines.
4
Solve for unknowns. Substitute, simplify, and use inverse trig functions for angles.
5
Verify. Check that angles sum to \(180^\circ\) and that the area formula \(\tfrac{1}{2}ab\sin C\) gives a consistent result.

Example: Triangle Area

Find the area of \(\triangle ABC\) with \(a = 10\), \(b = 7\), and included angle \(C = 40^\circ\).

\(40^\circ\)
\(a = 10\)
\(b = 7\)
A
B
C

Example: Law of Sines (AAS)

In \(\triangle ABC\), \(A = 50^\circ\), \(B = 70^\circ\), and \(a = 12\). Find side \(b\).

1
Find \(C\): \( C = 180^\circ - 50^\circ - 70^\circ = 60^\circ \)
2
Set up the proportion: \( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} \)
3
Substitute: \( \dfrac{12}{\sin 50^\circ} = \dfrac{b}{\sin 70^\circ} \)
4
Solve: \( b = \dfrac{12 \cdot \sin 70^\circ}{\sin 50^\circ} = \dfrac{12 \times 0.9397}{0.7660} \approx \mathbf{14.72} \)

Example: Law of Cosines (SAS)

In \(\triangle PQR\), \(p = 8\), \(q = 11\), and \(R = 130^\circ\). Find side \(r\).

\(130^\circ\)
\(p=8\)
\(q=11\)
\(r=?\)
P
R
Q

Example: Finding an Angle (SSS)

A triangle has sides \(a = 5\), \(b = 7\), \(c = 9\). Find the largest angle.

1
The largest angle is opposite the longest side. The longest side is \(c = 9\), so we find angle \(C\).
2
Rearrange: \( \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{25 + 49 - 81}{2(5)(7)} \)
3
Compute: \( \cos C = \dfrac{-7}{70} = -0.1 \)
4
Solve: \( C = \cos^{-1}(-0.1) \approx \mathbf{95.7^\circ} \) (obtuse, as expected)

The Ambiguous Case (SSA)

When given two sides and an angle opposite one of them (SSA), the Law of Sines may give zero, one, or two valid triangles.

0 Triangles
\(\sin B > 1\): the side is too short to reach the opposite side. No solution.
1 Triangle
\(\sin B = 1\) (right triangle), or the supplement angle would make the angle sum exceed \(180^\circ\).
2 Triangles
Both \(B\) and \(180^\circ - B\) produce valid angle sums. Check both!

Common Mistakes to Avoid

✕ Wrong Angle in Area Formula
Using any angle instead of the included angle between the two chosen sides. The \(C\) in \(\tfrac{1}{2}ab\sin C\) must be the angle between sides \(a\) and \(b\).
✕ Calculator in Degrees vs. Radians
Computing \(\sin 50^\circ\) while the calculator is set to radians gives a completely wrong value. Always verify your mode.
✕ Forgetting the Ambiguous Case
With SSA, failing to check whether \(180^\circ - B\) also gives a valid triangle. You may miss a second solution.

Formula Summary

Formula Expression Use When
Triangle Area \(\tfrac{1}{2}ab\sin C\) SAS (two sides + included angle)
Law of Sines \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) AAS, ASA, SSA
Law of Cosines \(c^2 = a^2 + b^2 - 2ab\cos C\) SAS, SSS
Inverse Cosine \(\cos C = \dfrac{a^2+b^2-c^2}{2ab}\) SSS (find any angle)

Question 1 of 5

What is the area of a triangle with sides \(a = 6\), \(b = 10\), and included angle \(C = 30^\circ\)?

Question 2 of 5

In \(\triangle ABC\), \(A = 45^\circ\), \(B = 60^\circ\), and \(a = 8\). Using the Law of Sines, find \(b\). Which value is closest?

Question 3 of 5

A triangle has sides \(a = 5\), \(b = 8\), and included angle \(C = 60^\circ\). What is \(c\)?

Question 4 of 5

Which formula setup is needed when you know all three sides of a triangle and want to find an angle?

Question 5 of 5

In the ambiguous case (SSA), you find \(\sin B = 0.85\). Angle \(A = 40^\circ\). Given these values, how many valid triangles exist?