Sine rule and cosine rule

Where trigonometric ratios are used to solve a right triangle, sine and cosine rules can solve any triangle.

The convention is that A^\hat A is the opposite angle to side aa.

Uppercase letter for angle is opposite to corresponding lowercase letter for side
Triangle naming convention

Contents

Sine rule

The sine rule states that the sides in increasing order correspond to their opposite angles in increasing order. Namely longest side is opposite to the largest angle; shortest side is opposite to the smallest angle.

Using the notation of triangles, it states

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

sin⁡\sin is positive for both acute and obtuse angles. The sine rule cannot distinguish between an acute or obtuse angle.

Note: The sine rule cannot be used on what may be the longest side of the triangle. Meanwhile, the second and third longest sides are guaranteed to be opposite to acute angles.

Tip: The angle opposite to the longest side should be found last using sum of angles in a triangle adding to 180°180\degree.

Example: Triangle ABCABC has AB=3AB = 3, BC=8BC = 8, AC=7AC = 7 and B^=60°\hat B = 60\degree. Find A^\hat A.

Because angle AA is opposite the longest side BCBC, we should instead find C^\hat C via sine rule.

bsin⁡B=csin⁡C7sin⁡60°=3sin⁡Csin⁡C=37⋅32C≈21.79°\begin{align*} \frac{b}{\sin B} &= \frac{c}{\sin C} \\ \frac{7}{\sin 60\degree} &= \frac{3}{\sin C} \\ \sin C &= \frac{3}{7} \cdot \frac{\sqrt 3}{2} \\ C &\approx 21.79\degree \\ \end{align*}

Now we can safely find the A^\hat A using triangle sum.

A≈180°−60°−21.79°≈98.2°■\begin{align*} A &\approx 180\degree - 60\degree - 21.79\degree \\ &\approx 98.2\degree \qed \end{align*}

Had we directly used sine rule on A^\hat A, we could not have discerned between 98.2°98.2\degree and 81.8°81.8\degree.

Cosine rule

The cosine rule is a generalization of our friend, Pythagorean theorem. It states

c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C

No order of side lengths of a,b,ca, b, c is presumed. The only condition is that angle CC is opposite to side cc.

Unlike the sine rule, the cosine rule can distinguish between acute and obtuse angles.

However, this can be more complicated to use.

Types of triangles

In addition to the classifications shown here, triangles can be classified by the three pieces of given information. Given all three angles are only sufficient to identify similar triangles. To identify the exact (congruent) triangle you need at least one side. ASA means the side between two angles are given; SAA means side outside the two angles are given.

typetrianglesfirst stepnote
SSS11cosine rulefirst find angle opposite to the longest side
SAS11cosine rule
SAA11find last angle
ASA11find last angle
SSA11 or 22sine rule“the ambiguous case”
AAA∞\inftysimilar triangles only

The SSA ambiguous case is discussed below. In all other cases after the first step, either all three sides or all three angles would be known. Continue using sine rule. The sine rule should not be used for the missing angle opposite to the longest side.

Generally speaking, the triangle you are being asked to solve, exists. Some triangles do not exist, such as when the triangle inequality is violated or it has two obtuse (or right) angles, so it is rather obvious when that happens. Every triple of sides that satisfies the triangle inequality forms a triangle.

SSA Ambiguous case

The ambiguous case is different in that usually two triangles exist. This is very different from sine rule not being able to discern which one of obtuse or acute. Ambiguous case is where both exist.

Draw height 'b sin A' with hypotenuse b
Number of triangles depends on how "a" compares to "b sin A".

Assume sides aa, bb and angle AA are given.

This shows the count of triangles in each case.

casecountdiagram
a<bsin⁡Aa < b\sin A0\quad 0SSA side 'a' too short
a=bsin⁡Aa = b\sin A1\quad 1SSA side 'a' same length as 'b sin A'
a∈]bsin⁡A,b[a \in \left]b\sin A, b\right[2\quad 2SSA two triangles
a≥ba \geq b1\quad 1SSA one triangle when 'a' is longer than 'b'

If not given in the question, it is recommended to confirm the number of triangles before continuing.

The ambiguous case with two triangles continues with an acute and an obtuse angle BB.

acute

  1. Find B1B_1 using sine rule, take the sin⁡−1\sin^{-1} value
  2. Sum of angles to find C1C_1
  3. Sine rule to find c1c_1

The same can be be done for obtuse, but there are some “shortcuts” if the acute triangle is already solved.

obtuse

  1. Find B2B_2 with B2=180°−B1B_2 = 180\degree - B_1
  2. Sum of angles to find C2C_2
  3. Sine rule to find c2c_2, or c2=c1−2acos⁡B1c_2 = c_1 - 2a\cos B_1

Applications

Solving a word problem requires drawing a detailed diagram. Here are some tips for doing so.

  1. Bearings are measured clockwise relative to North. They can be degrees or 3 digits representing the angle to the nearest degree.
  2. Is object 1 going at a certain direction, or is it seeing object 2 coming from a certain direction? Add 180°180\degree bearings for the second case.
  3. Is it an acute triangle or an obtuse triangle?
  4. distance = speed ×\times time