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

1. Hypotenuse:

• The hypotenuse is the side opposite the right angle. It is always the longest side of the triangle.

2. Adjacent Side:

• The adjacent side is the side next to the angle of interest, excluding the hypotenuse.

3. Opposite Side:

• The opposite side is the side opposite the angle of interest.

Pythagoras' Theorem is a fundamental relation in a right-angled triangle, stating: $\text{Hypotenuse}^2 = \text{Opposite Side}^2 + \text{Adjacent Side}^2$

Or in terms of the sides:

$c^2 = a^2 + b^2$

Where:

• $\ c$ is the length of the hypotenuse,
• $\ a \ and \ b$ are the lengths of the other two sides.

In a right-angled triangle, the primary trigonometric ratios are defined as follows:

• Sine (sin): $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$
• Cosine (cos): $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
• Tangent (tan): $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}$

Example Problem:

Problem: In a right-angled triangle, the length of the hypotenuse is 10 units, and one of the angles is $\ 30^\circ$. Find the lengths of the opposite and adjacent sides.

Solution:

4. Identify the Sides:

• Hypotenuse $\ c = 10$ units.
• The angle $\ \theta = 30^\circ .$
• We need to find the opposite side $\ a$ and adjacent side $\ b .$

5. Use the Sine Function to Find the Opposite Side: $\sin 30^\circ = \frac{a}{10}$ Since $\ \sin 30^\circ = \frac{1}{2} ,$ we have: $\frac{1}{2} = \frac{a}{10}$ Solving for $\ a$: $a = \frac{10}{2} = 5 \text{ units}$
6. Use the Cosine Function to Find the Adjacent Side: $\cos 30^\circ = \frac{b}{10}$ Since $\ \cos 30^\circ = \frac{\sqrt{3}}{2}$, we have: $\ \frac{\sqrt{3}}{2} = \frac{b}{10}$ Solving for $\ b :$ $b = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ units}$ Final Answer:

• Opposite side $\ a = :success[5 units]$.
• Adjacent side $\ b = :success[5\sqrt{3} units]$.

Example Problem for Area:

Problem: Find the area of a right-angled triangle where the lengths of the two shorter sides are 6 units and 8 units.

Solution:

7. Identify the Base and Height:

• Base $\ = 6$ units.
• Height $\ = 8$ units.

8. Calculate the Area: $\text{Area} = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 48 = 24 \text{ square units}$ Final Answer:

• The area is 24 square units.