Question 9

Right-Angled Triangle: With right angle at B, for angles A and C:

• Opposite and Adjacent swap between A and C — what is opposite for A is adjacent for C, and vice versa.

$\sin A = \frac{\text{Opp}_A}{\text{Hyp}}, \quad \cos A = \frac{\text{Adj}_A}{\text{Hyp}}, \quad \tan A = \frac{\text{Opp}_A}{\text{Adj}_A}$

Pythagoras Theorem: $AC^2 = AB^2 + BC^2$ — finds the hypotenuse once both legs are known.

We will use trigonometric ratios and the Pythagoras theorem.

Step 1 — Find the sides of the triangle

Let's draw a right-angled triangle. Angle B is the right angle. We are given $\tan A = \frac{1}{\sqrt{3}}$. We know that $\tan A = \frac{\text{Opposite side}}{\text{Adjacent side}}$. So, $\frac{BC}{AB} = \frac{1}{\sqrt{3}}$. Let $BC = \mathbf{k}$ and $AB = \mathbf{\sqrt{3}k}$. This is for some positive number $k$. Now, we use the Pythagoras theorem.

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

$AC^2 = (\sqrt{3}k)^2 + (k)^2$

$AC^2 = 3k^2 + k^2$

$AC^2 = 4k^2$

$AC = \sqrt{4k^2}$

$\boxed{AC = 2k}$

Step 2 — Calculate the required trigonometric ratios

Now we find the values of $\sin A$, $\cos A$, $\sin C$, and $\cos C$. For angle A, we have:

$\sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2}$

$\cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2}$

For angle C, we know that A and C are complementary angles. This means $C = 90^\circ - A$. So, $\sin C = \cos A$.

$\sin C = \frac{\sqrt{3}}{2}$

And $\cos C = \sin A$.

$\cos C = \frac{1}{2}$

Step 3 — Evaluate the first expression

Let's find the value of $\sin A \cos C + \cos A \sin C$. We substitute the ratios we found.

$\sin A \cos C + \cos A \sin C = \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$

$= \frac{1}{4} + \frac{3}{4}$

$= \frac{1+3}{4}$

$= \frac{4}{4}$

$\boxed{1}$

Step 4 — Evaluate the second expression

Now, let's find the value of $\cos A \cos C - \sin A \sin C$. We substitute the ratios again.

$\cos A \cos C - \sin A \sin C = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right) - \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right)$

$= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}$

$\boxed{0}$

Answer

(i) $1$ (ii) $0$