Trigonometry Beyond Right Triangles · Lesson 4

Law of sines and law of cosines

SOH-CAH-TOA needs a right angle. For any triangle, name the angles AA, BB, and CC and the sides across from them aa, bb, and cc. Two laws connect them. In this course the stem gives the formula; your job is to choose the right law and match each side with the angle across from it.

Law of sines: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}. Use it when you know an angle and the side across from it, plus one more side or angle. With two angles, find the third from the 180∘180^\circ sum first.

Law of cosines: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C. Use it with two sides and the angle between them (SAS) to find the third side, or with all three sides (SSS) to find an angle. When C=90∘C = 90^\circ, cos⁡C=0\cos C = 0 and it becomes the Pythagorean theorem.

Which law fits
You knowUseFirst step
Two sides and the angle between them (SAS)Law of cosinesFind the third side: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C
All three sides (SSS)Law of cosinesSolve for cos⁡C\cos C, with CC across from the side you put on the left
Two angles and a side (AAS or ASA)Law of sinesFind the third angle, then pair each side with its opposite angle

If the angle is obtuse, its cosine is negative, so −2abcos⁡C-2ab\cos C adds to a2+b2a^2 + b^2, and the third side is longer than the Pythagorean value. The largest angle is always across from the longest side.

Worked example Medium

Triangle ABC with two angles and one side marked1030°45°ABC
Triangle ABC with two angles and one side marked Triangle ABC with side AB along the bottom, A at the left end and B at the right end, and vertex C above the base, closer to B. The angle at A is marked 30° and the angle at B is marked 45°. Side BC, the side across from angle A, is labelled 10. Side AC, across from angle B, has no label.

In triangle ABC, the measure of angle A is 30∘30^\circ, the measure of angle B is 45∘45^\circ, and BC = 10. By the law of sines, asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}, where side aa is across from angle A and side bb is across from angle B. What is the length of AC‾\overline{AC}?

  1. 525\sqrt{2}
  2. 10210\sqrt{2}Answer
  3. 1515
  4. 2020

How to solve it

  1. AC is across from angle B, so it is bb. BC is across from angle A, so a=10a = 10.
  2. 10sin⁡30∘=bsin⁡45∘\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}.
  3. b=10sin⁡45∘sin⁡30∘=10⋅2212=102b = \frac{10 \sin 45^\circ}{\sin 30^\circ} = \frac{10 \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}} = 10\sqrt{2}.

Why each choice is right or wrong

  • A. Incorrect. 525\sqrt{2} is 10sin⁡30∘sin⁡45∘\frac{10\sin 30^\circ}{\sin 45^\circ}, the ratio flipped. AC sits across from the 45∘45^\circ angle, so sin⁡45∘\sin 45^\circ goes with AC: 10210\sqrt{2}.
  • B. Correct. 10sin⁡30∘=bsin⁡45∘\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}, so b=20⋅22=102b = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2}, about 14.1.
  • C. Incorrect. 1515 is 10⋅453010 \cdot \frac{45}{30}, a ratio of the angles themselves. The law of sines uses their sines, and sin⁡45∘sin⁡30∘=2\frac{\sin 45^\circ}{\sin 30^\circ} = \sqrt{2}, not 1.5.
  • D. Incorrect. 2020 is 10sin⁡30∘\frac{10}{\sin 30^\circ}, the shared ratio. That is one step short: multiply by sin⁡45∘\sin 45^\circ to get 10210\sqrt{2}.

Worked example Hard

Two sides of a triangular garden bed measure 6 feet and 10 feet, and the angle between them measures 120∘120^\circ. By the law of cosines, c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C. What is the length, in feet, of the third side?

  1. 2192\sqrt{19}
  2. 2342\sqrt{34}
  3. 1414Answer
  4. 196196

How to solve it

  1. The known angle is between the known sides (SAS), so use the law of cosines with C=120∘C = 120^\circ.
  2. cos⁡120∘=−12\cos 120^\circ = -\frac{1}{2} (reference angle 60∘60^\circ, quadrant II).
  3. c2=36+100−2(6)(10)(−12)=136+60=196c^2 = 36 + 100 - 2(6)(10)(-\frac{1}{2}) = 136 + 60 = 196.
  4. c=196=14c = \sqrt{196} = 14 feet.

Why each choice is right or wrong

  • A. Incorrect. 2192\sqrt{19} is 136−60\sqrt{136 - 60}, which uses cos⁡120∘=12\cos 120^\circ = \frac{1}{2}. An obtuse angle has a negative cosine, so −2abcos⁡C-2ab\cos C adds 60.
  • B. Incorrect. 2342\sqrt{34} is 36+100\sqrt{36 + 100}, the Pythagorean theorem. The angle is 120∘120^\circ, not 90∘90^\circ, so the −2abcos⁡C-2ab\cos C term stays.
  • C. Correct. c2=36+100−120cos⁡120∘=136+60=196c^2 = 36 + 100 - 120\cos 120^\circ = 136 + 60 = 196, so c=14c = 14.
  • D. Incorrect. 196196 is c2c^2. That is one step short: take the square root to get 14.

Trap Pairing a side with the wrong angle

In each fraction of the law of sines, the side and the angle sit across from each other. Mark the pairs on the figure before you write the ratio. A flipped ratio, or a side paired with a neighboring angle, gives a tempting wrong choice.

Trap The sign of $-2ab\cos C$

Keep the minus sign and the sign of cos⁡C\cos C together. For C=120∘C = 120^\circ, cos⁡C=−12\cos C = -\frac{1}{2}, so −2abcos⁡C=+ab-2ab\cos C = +ab. Dropping either sign makes the third side too short.

Calculator When a calculator isn’t faster

When the choices are exact, such as 10210\sqrt{2}, or written as expressions such as 12sin⁡40∘sin⁡65∘\frac{12\sin 40^\circ}{\sin 65^\circ}, setting up the law is the real work. When the choices are decimals, a calculator evaluates sin⁡\sin and cos⁡\cos quickly.

Check the angle mode first: degree/radian mode is the classic calculator trap. Triangle angles are in degrees, so use DEG on a handheld, or choose Degrees in the wrench menu of the built-in Desmos calculator if you test online.