Non-Right-Angled Triangles (Edexcel A-Level Mathematics): Revision Notes

Example Using the Sine Rule:

Problem: In a triangle $\ ABC , \ \angle A = 45^\circ , \ \angle B = 60^\circ \, and \ a = 7$ $cm$. Find the length of side $\ b .$

Solution:

1. Find the third angle $C$: $\angle C = 180^\circ - 45^\circ - 60^\circ = 75^\circ$
2. Apply the Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B}$ Substituting the known values: $\frac{7}{\sin 45^\circ} = \frac{b}{\sin 60^\circ}$ Solving for $b$: $b = \frac{7 \times \sin 60^\circ}{\sin 45^\circ} = \frac{7 \times \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 7 \times \frac{\sqrt{3}}{\sqrt{2}} \approx 8.57 \text{ cm}$ Final Answer:

• $\ b \approx :highlight[8.57]$ $cm$.

The Cosine Rule states that: $c^2 = a^2 + b^2 - 2ab \cos C$

Where:

• $\ c$ is the side opposite angle $\ C ,$
• $\ a \ and \ b$ are the other two sides. To find an angle, the Cosine Rule can be rearranged as:

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

Example Using the Cosine Rule:

Problem: In a triangle $\ ABC$, the sides $\ a = 5$ $cm$, $\ b = 7$ $cm$, and $\ c = 10$ $cm$. Find $\ \angle C .$

Solution:

3. Apply the Cosine Rule to find $\ \angle C :$ $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Substituting the known values: $\cos C = \frac{5^2 + 7^2 - 10^2}{2 \times 5 \times 7}$ $\cos C = \frac{25 + 49 - 100}{70} = \frac{-26}{70} \approx -0.3714$
4. Find $\ \angle C$ using the inverse cosine function: $\angle C = \cos^{-1}(-0.3714) \approx 111.5^\circ$ Final Answer:

• $\ \angle C \approx :highlight[111.5^\circ] .$

The area$\ A$ of a non-right-angled triangle can be calculated using the following formula, based on two sides and the included angle: $\text{Area} = \frac{1}{2}ab\sin C$

Where:

• $\ a \ and \ b$ are the lengths of two sides,
• $\ C$ is the angle between them.

Example for Area Calculation:

Problem: In a triangle$\ ABC \, \ a = 8 \ cm, \ b = 6$ $cm$, and $\ \angle C = 30^\circ$. Find the area of the triangle.

Solution:

5. Apply the area formula: $\text{Area} = \frac{1}{2} \times 8 \times 6 \times \sin 30^\circ$ Since$\ \sin 30^\circ = \frac{1}{2} :$ $\text{Area} = \frac{1}{2} \times 8 \times 6 \times \frac{1}{2} = \frac{1}{2} \times 24 = 12 \text{ square cm}$ Final Answer:

• The area of the triangle is $\ :highlight[12]$ $square$ $cm$.