The Sine Rule (VCE SSCE General Mathematics): Revision Notes

Example: Finding an Angle Using the Sine Rule

Find angle $B$ in the triangle shown, where angle $A = 120°$, side $a = 7$ and side $b = 6$. Give your answer to one decimal place.

Solution:

We have the angle-side pairs: $a = 7$ with $A = 120°$, and $b = 6$ with $B = ?$

Using the sine rule:

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

Substituting the known values:

$\frac{7}{\sin 120°} = \frac{6}{\sin B}$

Cross-multiplying:

$7 \times \sin B = 6 \times \sin 120°$

Dividing both sides by $7$:

$\sin B = \frac{6 \times \sin 120°}{7}$

Finding angle $B$:

$B = \sin^{-1}\left(\frac{6 \times \sin 120°}{7}\right)$

Using a calculator:

$B = 47.928...°$

Therefore, angle $B = 47.9°$ (to one decimal place).

Example: Finding a Side Using the Sine Rule

Find side $c$ in a triangle where angle $A = 100°$, angle $C = 50°$, and side $b = 8$. Give your answer to one decimal place.

Solution:

First, find angle $B$ using the angle sum property:

$A + B + C = 180°$

$100° + B + 50° = 180°$

$B + 150° = 180°$

$B = 30°$

Now we have the angle-side pairs: $b = 8$ with $B = 30°$, and $c = ?$ with $C = 50°$

Using the sine rule:

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

Substituting the known values:

$\frac{8}{\sin 30°} = \frac{c}{\sin 50°}$

Multiplying both sides by $\sin 50°$:

$c = \frac{8 \times \sin 50°}{\sin 30°}$

Using a calculator:

$c = 12.256...$

Therefore, side $c = 12.3$ units (to one decimal place).

Example: Solving the Ambiguous Case

In triangle $ABC$, angle $A = 40°$, side $c = 9$ cm, and side $a = 6$ cm.

When we draw side $c$ from vertex $B$ at $40°$ to the base, and then measure out side $a = 6$ cm from vertex $B$, this side can cross the base in two possible places, which we call $C$ and $C'$.

This creates two possible triangles: triangle $ABC$ (shown in blue) and triangle $ABC'$ (shown in red).

To find both possible values for angle $C$:

Step 1: Use the sine rule to find one value of angle $C$:

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

$\frac{\sin C}{9} = \frac{\sin 40°}{6}$

$\sin C = \frac{9 \times \sin 40°}{6}$

$C = \sin^{-1}\left(\frac{9 \times \sin 40°}{6}\right)$

$C = 74.62°$

Step 2: Identify which angle this represents. Since $\angle BC'A$ is clearly greater than $90°$, this $74.62°$ must be $\angle BCA$.

Step 3: Triangle $C'BC$ is isosceles (both marked sides equal $6$ cm), so the base angles are equal:

$\angle BC'C = \angle BCC' = 74.62°$

Step 4: Angles on a straight line sum to $180°$:

$\angle BC'A + 74.62° = 180°$

$\angle BC'A = 180° - 74.62° = 105.38°$

Therefore, the two possible values for angle $C$ are 74.62° and 105.38°.

Example: Navigation Problem

Leo wants to tie a rope from a tree at point $A$ to a tree at point $B$ on the other side of a river. When he stood at $A$, he saw tree $B$ at an angle of $50°$ with the riverbank. After walking $200$ metres east to point $C$, the tree was at an angle of $30°$ with the riverbank.

Find the length of rope required to reach from $A$ to $B$ to two decimal places.

Solution:

The unknown is the length of rope, which is side $c$. We know angle $C = 30°$ and side $b = 200$ m.

We need to find angle $B$ first. Using the angle sum:

$A + B + C = 180°$

$50° + B + 30° = 180°$

$B = 100°$

Now we have: $b = 200$ m with $B = 100°$, and $c = ?$ with $C = 30°$

Using the sine rule:

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

$\frac{c}{\sin 30°} = \frac{200}{\sin 100°}$

Multiplying both sides by $\sin 30°$:

$c = \frac{200 \times \sin 30°}{\sin 100°}$

$c = 101.542...$

Therefore, the rope must be 101.54 m long (to two decimal places).