Sine and cosine rules

Sine and Cosine Rules

Introduction to Sine and Cosine Rules

The Sine and Cosine Rules are fundamental in trigonometry for solving triangles when we do not have a right-angled triangle. These rules relate the lengths of the sides of a triangle to the sines and cosines of its angles.

Sine Rule

Consider a triangle $\triangle ABC$ with sides $a$, $b$, and $c$ opposite to angles $A$, $B$, and $C$ respectively. The Sine Rule states:

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

This relationship is particularly useful when we know two angles and one side, or two sides and a non-included angle.

Derivation of the Sine Rule

Let's derive the Sine Rule using the area of a triangle.

The area of $\triangle ABC$ can be expressed in terms of two sides and the sine of the included angle:

$\text{Area of } \triangle ABC = \frac{1}{2} \times a \times b \times \sin C$

$\text{Area of } \triangle ABC = \frac{1}{2} \times a \times c \times \sin B$

$\text{Area of } \triangle ABC = \frac{1}{2} \times b \times c \times \sin A$

Equating these expressions, we get:

$\frac{1}{2} \times a \times b \times \sin C = \frac{1}{2} \times a \times c \times \sin B = \frac{1}{2} \times b \times c \times \sin A$

Dividing each term by $\frac{1}{2} \times a \times c$, we have:

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

This is the Sine Rule.

Example 1

Find the unknown side and angle in triangle $\triangle ABC$ given:

• Side $a = 7.5$ cm
• Side $c = 8.6$ cm
• Angle $C = 80^\circ$

Solution

Using the Sine Rule:

$\frac{a}{\sin A} = \frac{c}{\sin C}$

Substitute the known values:

$\frac{7.5}{\sin A} = \frac{8.6}{\sin 80^\circ}$

First, calculate $\sin 80^\circ$:

$\sin 80^\circ \approx 0.9848$

Now solve for $\sin A$:

$\frac{7.5}{\sin A} = \frac{8.6}{0.9848}$

$\sin A = \frac{7.5 \times 0.9848}{8.6} \approx 0.8597$

Thus, $A \approx \arcsin(0.8597) \approx 59.5^\circ$.

To find angle $B$, use the fact that the sum of angles in a triangle is $180^\circ$:

$B = 180^\circ - A - C = 180^\circ - 59.5^\circ - 80^\circ = 40.5^\circ$

Finally, use the Sine Rule to find side $b$:

$\frac{b}{\sin B} = \frac{7.5}{\sin 59.5^\circ}$

$b = \frac{7.5 \times \sin 40.5^\circ}{\sin 59.5^\circ} \approx \frac{7.5 \times 0.6505}{0.8597} \approx 5.7 \text{ cm}$

Therefore, the unknown side $b \approx 5.7$ cm and angle $A \approx 59.5^\circ$.

Example 2

Find the unknown sides and angle in triangle $\triangle ABC$ where:

• Side $a = 22.2$ cm
• Angle $B = 86^\circ$
• Angle $A = 26^\circ$

Solution

First, find angle $C$:

$C = 180^\circ - A - B = 180^\circ - 26^\circ - 86^\circ = 68^\circ$

Using the Sine Rule to find side $b$:

$\frac{b}{\sin B} = \frac{a}{\sin A}$

$b = \frac{22.2 \times \sin 86^\circ}{\sin 26^\circ}$

$\sin 86^\circ \approx 0.9976, \quad \sin 26^\circ \approx 0.4384$

$b \approx \frac{22.2 \times 0.9976}{0.4384} \approx 50.5 \text{ cm}$

Using the Sine Rule to find side $c$:

$\frac{c}{\sin C} = \frac{a}{\sin A}$

$c = \frac{22.2 \times \sin 68^\circ}{\sin 26^\circ}$

$\sin 68^\circ \approx 0.9272$

$c \approx \frac{22.2 \times 0.9272}{0.4384} \approx 46.9 \text{ cm}$

Example 3

Find the unknown sides and angles in triangle $\triangle ABC$ where:

• Side $a = 3$ cm
• Side $c = 4$ cm
• Angle $B = 30^\circ$

Solution

By the Cosine Rule:

Consider a triangle $\triangle ABC$ with coordinates $A(0, 0)$, $B(c, 0)$ and $C(b \cos A, b \sin A)$.

$\cos A = \frac{x}{b}, \quad \sin A = \frac{y}{b}$

$x = b \cos A, \quad y = b \sin A$

Using the distance formula:

$a^2 = (b \cos A - c)^2 + (b \sin A - 0)^2$

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

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

Since $\sin^2 A + \cos^2 A = 1$:

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

Similarly, the Cosine Rule can be written for the other sides:

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

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