Sine and Cosine Rule

Introduction

In any triangle, the sides and angles are connected by certain fundamental relationships. The most important of these are the Sine Rule and the Cosine Rule, which allow us to solve for unknown sides and angles.

In a triangle ABC, the angles are denoted by capital letters A, B, C, and the lengths of the sides opposite to these angles are denoted by small letters a, b, c, respectively.

The Sine Rule (or Law of Sines)

The Sine Rule states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant for all three sides and is equal to the diameter of the triangle's circumcircle (2R).

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

where R is the circumradius of the triangle.

• When to use: The Sine Rule is useful when you know:

  1. Two angles and one side (AAS or ASA).

  2. Two sides and a non-included angle (SSA - this is the ambiguous case and may yield 0, 1, or 2 possible triangles).

The Cosine Rule (or Law of Cosines)

The Cosine Rule relates the lengths of the three sides of a triangle to the cosine of one of its angles. It is a generalization of the Pythagorean theorem.

$$a^2 = b^2 + c^2 - 2bc\cos A$$

$$b^2 = a^2 + c^2 - 2ac\cos B$$

$$c^2 = a^2 + b^2 - 2ab\cos C$$

The formulas can be rearranged to find an angle if all three sides are known:

$$\cos A = \frac{b^2+c^2-a^2}{2bc}$$

• When to use: The Cosine Rule is useful when you know:

  1. Two sides and the included angle (SAS).

  2. All three sides (SSS).

Example 1: Using Sine Rule (AAS)

Question: In a triangle ABC, if $A=30^\circ, B=45^\circ$ and side $a=10$, find side b.

Solution:

Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B}$.

$$\frac{10}{\sin 30^\circ} = \frac{b}{\sin 45^\circ}$$

$$b = \frac{10 \cdot \sin 45^\circ}{\sin 30^\circ} = \frac{10 \cdot (1/\sqrt{2})}{1/2} = \frac{20}{\sqrt{2}} = 10\sqrt{2}$$

Example 2: Using Cosine Rule (SAS)

Question: In a triangle ABC, sides $a=4, b=5$ and angle $C=60^\circ$. Find the length of side c.

Solution:

Using the Cosine Rule, $c^2 = a^2 + b^2 - 2ab\cos C$.

$$c^2 = 4^2 + 5^2 - 2(4)(5)\cos 60^\circ$$

$$c^2 = 16 + 25 - 40(1/2) = 41 - 20 = 21$$

$$c = \sqrt{21}$$

Example 3: Using Cosine Rule (SSS)

Question: The sides of a triangle are a=7, b=8, c=5. Find the angle A.

Solution:

Using the Cosine Rule for angles: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.

$$\cos A = \frac{8^2+5^2-7^2}{2(8)(5)} = \frac{64+25-49}{80} = \frac{40}{80} = \frac{1}{2}$$

Since $\cos A = 1/2$, the angle $A = 60^\circ$.

Example 4: Finding the Circumradius

Question: In $\triangle ABC$, if $a=4$ and $A=30^\circ$, find the circumradius R.

Solution:

From the Sine Rule, $\frac{a}{\sin A} = 2R$.

$$2R = \frac{4}{\sin 30^\circ} = \frac{4}{1/2} = 8$$

So, the circumradius $R=4$.

Example 5: Proving an Identity with Sine Rule

Question: In any triangle ABC, prove that $a(\sin B - \sin C) + b(\sin C - \sin A) + c(\sin A - \sin B) = 0$.

Solution:

From the Sine Rule, we have $a=2R\sin A, b=2R\sin B, c=2R\sin C$. Substitute these into the expression:

LHS = $2R\sin A(\sin B - \sin C) + 2R\sin B(\sin C - \sin A) + 2R\sin C(\sin A - \sin B)$

Factor out 2R and expand:

$= 2R[\sin A\sin B - \sin A\sin C + \sin B\sin C - \sin B\sin A + \sin C\sin A - \sin C\sin B]$

All the terms cancel out, so the result is $2R[0] = 0$ = RHS.

Example 6: Proving an Identity with Cosine Rule (Projection Formula)

Question: In any triangle ABC, prove that $b\cos C + c\cos B = a$.

Solution:

This is the Projection Formula. We can prove it using the Cosine Rule.

$$\cos C = \frac{a^2+b^2-c^2}{2ab} \quad \text{and} \quad \cos B = \frac{a^2+c^2-b^2}{2ac}$$

LHS = $b\left(\frac{a^2+b^2-c^2}{2ab}\right) + c\left(\frac{a^2+c^2-b^2}{2ac}\right) = \frac{a^2+b^2-c^2}{2a} + \frac{a^2+c^2-b^2}{2a}$

$$= \frac{a^2+b^2-c^2+a^2+c^2-b^2}{2a} = \frac{2a^2}{2a} = a = \text{RHS}$$

Example 7: Solving a Triangle

Question: In a triangle ABC, $A=45^\circ, B=60^\circ, a=2$. Find the remaining angle and sides.

Solution:

• Angle C: The sum of angles in a triangle is $180^\circ$. So, $C = 180^\circ - (A+B) = 180^\circ - (45^\circ+60^\circ) = 75^\circ$.

• Side b: Using Sine Rule, $\frac{b}{\sin B} = \frac{a}{\sin A} \implies \frac{b}{\sin 60^\circ} = \frac{2}{\sin 45^\circ} \implies b = \frac{2\sin 60^\circ}{\sin 45^\circ} = \frac{2(\sqrt{3}/2)}{1/\sqrt{2}} = \sqrt{6}$.

• Side c: Using Sine Rule, $\frac{c}{\sin C} = \frac{a}{\sin A} \implies \frac{c}{\sin 75^\circ} = \frac{2}{\sin 45^\circ}$. We know $\sin 75^\circ = \frac{\sqrt{3}+1}{2\sqrt{2}}$.

  $c = \frac{2\sin 75^\circ}{\sin 45^\circ} = \frac{2((\sqrt{3}+1)/2\sqrt{2})}{1/\sqrt{2}} = \sqrt{3}+1$.

Example 8: Ratio of Sides and Angles

Question: In a $\triangle ABC$, if the sides are in the ratio $1:\sqrt{3}:2$, find the ratio of the angles.

Solution:

Let the sides be $a=k, b=k\sqrt{3}, c=2k$. We find the angles using the Cosine Rule.

$$\cos A = \frac{(k\sqrt{3})^2+(2k)^2-k^2}{2(k\sqrt{3})(2k)} = \frac{3k^2+4k^2-k^2}{4k^2\sqrt{3}} = \frac{6k^2}{4k^2\sqrt{3}} = \frac{\sqrt{3}}{2} \implies A=30^\circ$$

$$\cos B = \frac{k^2+(2k)^2-(k\sqrt{3})^2}{2(k)(2k)} = \frac{k^2+4k^2-3k^2}{4k^2} = \frac{2k^2}{4k^2} = \frac{1}{2} \implies B=60^\circ$$

Then $C = 180^\circ - (30^\circ+60^\circ) = 90^\circ$. The ratio of angles A:B:C is $30:60:90$ or $1:2:3$.

Example 9: Ambiguous Case (SSA)

Question: In a triangle, $a=4, b=5, A=30^\circ$. Find the possible values for angle B.

Solution:

Using Sine Rule: $\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{b\sin A}{a} = \frac{5\sin 30^\circ}{4} = \frac{5(1/2)}{4} = \frac{5}{8}$.

Since $0 < \sin B < 1$, there are two possible angles for B in the range $(0, 180^\circ)$: an acute angle $B_1 = \arcsin(5/8)$ and an obtuse angle $B_2 = 180^\circ - \arcsin(5/8)$. Since $A=30^\circ$, both $A+B_1 < 180^\circ$ and $A+B_2 < 180^\circ$ are possible. Thus, there are two possible triangles.

Example 10: Finding Sides from Angles and Perimeter

Question: The angles of a triangle are in the ratio 1:2:3. If the perimeter is $3+\sqrt{3}$, find the sides.

Solution:

Let the angles be $x, 2x, 3x$. The sum of angles is $x+2x+3x = 6x = 180^\circ \implies x=30^\circ$.

The angles are $A=30^\circ, B=60^\circ, C=90^\circ$.

By the Sine Rule, the ratio of the sides is $a:b:c = \sin A : \sin B : \sin C = \sin 30^\circ : \sin 60^\circ : \sin 90^\circ = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$.

This simplifies to the ratio $1:\sqrt{3}:2$. Let the sides be $k, k\sqrt{3}, 2k$.

Perimeter = $k + k\sqrt{3} + 2k = k(3+\sqrt{3})$.

We are given the perimeter is $3+\sqrt{3}$.

$k(3+\sqrt{3}) = 3+\sqrt{3} \implies k=1$.

So, the sides are $1, \sqrt{3}, 2$.