Law of Sines

The Pythagorean theorem and SOH-CAH-TOA only work for right triangles. But most triangles in the real world are not right triangles. The Law of Sines lets you solve any triangle, as long as you know enough information.

The Formula

sin(A) / a = sin(B) / b = sin(C) / c

Here A, B, C are the angles and a, b, c are the sides opposite those angles respectively. The law says: the ratio of the sine of an angle to its opposite side is the same for all three angle-side pairs.

When to Use the Law of Sines

Given Information	Case Name	Use Law of Sines?
Two angles and a side between them	ASA	Yes (find 3rd angle first)
Two angles and a non-included side	AAS	Yes
Two sides and an angle opposite one	SSA	Yes (ambiguous case -- careful!)
Two sides and the included angle	SAS	No -- use Law of Cosines

Worked Example 1: AAS Case

In triangle ABC, A = 40 degrees, B = 70 degrees, a = 10. Find side b.

Find angle C: C = 180 − 40 − 70 = 70 degrees.
Apply Law of Sines: sin(40)/10 = sin(70)/b
Solve: b = 10 · sin(70) / sin(40) = 10 · 0.9397 / 0.6428 = 14.62

The Ambiguous Case (SSA)

When you know two sides and an angle opposite one of them (SSA), there may be zero, one, or two possible triangles. This is called the ambiguous case.

Worked Example 2: SSA -- Two Solutions

Given: a = 8, b = 12, A = 30 degrees. Find angle B.

sin(30)/8 = sin(B)/12

sin(B) = 12 · sin(30)/8 = 12 · 0.5/8 = 0.75

B = arcsin(0.75) = 48.59 degrees

But wait -- sine is also positive in quadrant II: B could also be 180 − 48.59 = 131.41 degrees.

Check: A + B = 30 + 131.41 = 161.41 < 180. This works too!

Two valid triangles: one with B = 48.59 degrees, another with B = 131.41 degrees.

Triangle Area Formula

The Law of Sines gives us a convenient area formula for any triangle:

Area = (1/2) \cdot a \cdot b \cdot sin(C)

where a and b are two sides and C is the included angle between them.

Worked Example 3: Area

Find the area of a triangle with sides a = 7, b = 10, and included angle C = 55 degrees.

Area = (1/2)(7)(10) sin(55) = 35 · 0.8192 = 28.67 square units

Common Mistake

In the ambiguous case, students often forget to check whether the second possible angle (180 − B) also forms a valid triangle. Always verify that the sum of all angles is less than 180 degrees before rejecting or accepting a solution.

Matching Pairs

Each ratio in the Law of Sines links an angle with its opposite side. Always double-check that you are pairing each angle with the side across from it in the triangle -- not the side adjacent to it.

Practice Problems

1. In triangle ABC, A = 50 degrees, B = 60 degrees, a = 15. Find b.

Solution

sin(50)/15 = sin(60)/b. So b = 15 sin(60)/sin(50) = 15(0.8660)/0.7660 = 16.96.

2. In triangle ABC, A = 35 degrees, C = 85 degrees, c = 20. Find a.

Solution

B = 180 − 35 − 85 = 60 degrees. sin(35)/a = sin(85)/20. So a = 20 sin(35)/sin(85) = 20(0.5736)/0.9962 = 11.52.

3. Given a = 10, b = 14, A = 42 degrees. How many triangles are possible?

Solution

sin(B) = 14 sin(42)/10 = 14(0.6691)/10 = 0.9368. B = arcsin(0.9368) = 69.5 degrees. Second possibility: B = 110.5 degrees. Check: 42 + 110.5 = 152.5 < 180. Both work. Two triangles are possible.

4. Find the area of a triangle with sides 8 and 11, with an included angle of 40 degrees.

Solution

Area = (1/2)(8)(11) sin(40) = 44(0.6428) = 28.28 square units.

5. In triangle ABC, a = 5, A = 30 degrees, b = 20. How many triangles are possible?

Solution

sin(B) = 20 sin(30)/5 = 20(0.5)/5 = 2. Since sin(B) cannot exceed 1, no triangle is possible.

Summary

Law of Sines: sin(A)/a = sin(B)/b = sin(C)/c.
Use for AAS, ASA, and SSA configurations.
The SSA case is ambiguous -- check for 0, 1, or 2 solutions.
Area = (1/2)ab sin(C) for any triangle using two sides and the included angle.