The Sine Rule (Leaving Cert Mathematics): Revision Notes

Worked Example: Finding a Side Length

Let's find the length of side $x$ in a triangle where we know one side is 12 units, with angles of 34° and 62°.

Step 1: Set up the Sine Rule with sides on top (since we're finding a side) $\frac{x}{\sin 62°} = \frac{12}{\sin 34°}$

Step 2: Calculate the sine values

• $\sin 62° = 0.8829$
• $\sin 34° = 0.5592$

Step 3: Substitute and solve $\frac{x}{0.8829} = \frac{12}{0.5592}$

Step 4: Cross multiply $x \times 0.5592 = 12 \times 0.8829$ $x \times 0.5592 = 10.5948$

Step 5: Solve for $x$ $x = \frac{10.5948}{0.5592} = 18.946$

Answer: $x = 19$ units (to the nearest whole number)

Worked Example: Finding an Angle and Calculating Area

Let's find angle $A$ in a triangle where side $BC = 12$ cm, side $AC = 14$ cm, and angle $C = 53°$.

Step 1: Set up the Sine Rule with angles on top (since we're finding an angle) $\frac{\sin 53°}{\sin A} = \frac{12}{14}$

Step 2: Rearrange to solve for $\sin A$ $\frac{12}{\sin 53°} = \frac{14}{\sin A}$

Step 3: Calculate $\sin 53° = 0.7986$ $\frac{12}{0.7986} = \frac{14}{\sin A}$

Step 4: Cross multiply and solve $12 \sin A = 14 \times 0.7986$ $12 \sin A = 11.1804$ $\sin A = \frac{11.1804}{12} = 0.9317$

Step 5: Find angle $A$ using inverse sine $A = \sin^{-1}(0.9317) = 68.70°$

Answer: $A = :success[69°]$ (to the nearest degree)

Finding the area: Now we can calculate the triangle's area. First, find angle $B$: $B = 180° - 53° - 69° = 58°$

Using the area formula: Area = $\frac{1}{2}ab\sin C$

Area = $\frac{1}{2} \times 12 \times 14 \times \sin 58°$

Area = $84 \times 0.8480 = 71.2 cm²$