Question 4

Right-Angled Triangle: For any angle A, the sides are:

• Opposite — side facing angle A
• Adjacent — side next to angle A
• Hypotenuse — longest side, opposite the 90° angle

The relevant ratios here:

$\cot A = \frac{\text{Adjacent}}{\text{Opposite}}, \quad \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \sec A = \frac{\text{Hypotenuse}}{\text{Adjacent}}$

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

Since $\cot A = \frac{8}{15}$, we take adjacent $= 8k$ and opposite $= 15k$, then find the hypotenuse using Pythagoras.

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

Step 1 — Find the hypotenuse

We are given the value of $15 \cot A$.

Let's write $\cot A$ in a simpler form.

$15 \cot A = 8$

$\cot A = \frac{8}{15}$

We know that $\cot A$ is the ratio of the adjacent side to the opposite side.

Let the adjacent side be AB and the opposite side be BC.

So, let $AB = 8k$ and $BC = 15k$ for some positive number $k$.

Now, we use the Pythagoras theorem to find the hypotenuse AC.

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

$AC^2 = (8k)^2 + (15k)^2$

$AC^2 = 64k^2 + 225k^2$

$AC^2 = 289k^2$

$AC = \sqrt{289k^2}$

$\boxed{AC = 17k}$

Step 2 — Find sin A and sec A

Now we can find $\sin A$ using the sides of the triangle.

$\sin A$ is the ratio of the opposite side to the hypotenuse.

$\sin A = \frac{BC}{AC}$

$\sin A = \frac{15k}{17k}$

$\sin A = \frac{15}{17}$

Next, let's find $\sec A$.

$\sec A$ is the ratio of the hypotenuse to the adjacent side.

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

$\sec A = \frac{17k}{8k}$

$\sec A = \frac{17}{8}$

Answer

(i) $\sin A = \frac{15}{17}$ (ii) $\sec A = \frac{17}{8}$