Question 3

Right-Angled Triangle: A triangle with one angle equal to 90°. For any angle A in a right-angled triangle:

• Opposite — the side facing angle A
• Adjacent — the side next to angle A (not the hypotenuse)
• Hypotenuse — the longest side, opposite the 90° angle

The trig ratios are defined as:

$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \tan A = \frac{\text{Opposite}}{\text{Adjacent}}$

Pythagoras Theorem: $\text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2$

Since $\sin A = \frac{3}{4}$, we take opposite $= 3k$ and hypotenuse $= 4k$, then find adjacent using Pythagoras.

We will use a right-angled triangle and the Pythagoras theorem.

Step 1 — Find the missing side

Let's draw a right-angled triangle. We know that $\sin A = \frac{\text{Opposite side}}{\text{Hypotenuse}}$. Given $\sin A = \frac{3}{4}$. So, the opposite side is $\mathbf{3k}$. The hypotenuse is $\mathbf{4k}$. Let's use the Pythagoras theorem. The square of the hypotenuse equals the sum of the squares of the other two sides. Let the triangle be ABC, right-angled at B. Then BC is the opposite side to angle A. AC is the hypotenuse. AB is the adjacent side.

$AC^2 = AB^2 + BC^2$

$(4k)^2 = AB^2 + (3k)^2$

$16k^2 = AB^2 + 9k^2$

$AB^2 = 16k^2 - 9k^2$

$AB^2 = 7k^2$

$AB = \sqrt{7k^2}$

$\boxed{AB = \sqrt{7}k}$

Step 2 — Calculate cos A

We know that $\cos A = \frac{\text{Adjacent side}}{\text{Hypotenuse}}$. Let's substitute the side lengths we found.

$\cos A = \frac{AB}{AC}$

$\cos A = \frac{\sqrt{7}k}{4k}$

$\boxed{\cos A = \frac{\sqrt{7}}{4}}$

Step 3 — Calculate tan A

We know that $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}}$. Let's substitute the side lengths we found.

$\tan A = \frac{BC}{AB}$

$\tan A = \frac{3k}{\sqrt{7}k}$

$\boxed{\tan A = \frac{3}{\sqrt{7}}}$

Answer

(i) $\cos A = \frac{\sqrt{7}}{4}$ (ii) $\tan A = \frac{3}{\sqrt{7}}$