Question 10

Right-Angled Triangle: With right angle at Q, PR is the hypotenuse (side opposite the 90° angle). PQ and QR are the two legs.

Pythagoras Theorem: $\text{Hypotenuse}^2 = \text{Leg}_1^2 + \text{Leg}_2^2$, i.e. $PR^2 = PQ^2 + QR^2$

Trigonometric Ratios for angle P (opposite = QR, adjacent = PQ, hypotenuse = PR):

$\sin P = \frac{QR}{PR}, \quad \cos P = \frac{PQ}{PR}, \quad \tan P = \frac{QR}{PQ}$

We need to find the lengths of the triangle sides first.

Step 1 — Find side lengths

Let's draw the triangle. Angle Q is the right angle. We are given $\text{PQ} = \mathbf{5\text{ cm}}$. We know $\text{PR} + \text{QR} = \mathbf{25\text{ cm}}$. Let $\text{QR} = x$. Then $\text{PR} = \mathbf{25 - x}$. We use the Pythagoras theorem. It states that $\text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2$.

$\text{PR}^2 = \text{PQ}^2 + \text{QR}^2$

$(25 - x)^2 = (5)^2 + (x)^2$

$625 - 50x + x^2 = 25 + x^2$

$625 - 50x = 25$

$625 - 25 = 50x$

$600 = 50x$

$x = \frac{600}{50}$

$x = 12$

So, $\text{QR} = \mathbf{12\text{ cm}}$. Now we find PR.

$\text{PR} = 25 - x$

$\text{PR} = 25 - 12$

$\boxed{\text{PR} = 13\text{ cm}}$

Step 2 — Calculate trigonometric ratios

Now we have all side lengths. $\text{PQ} = \mathbf{5\text{ cm}}$ (adjacent to P). $\text{QR} = \mathbf{12\text{ cm}}$ (opposite to P). $\text{PR} = \mathbf{13\text{ cm}}$ (hypotenuse). Let's find $\sin P$.

$\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}}$

$\sin P = \frac{\text{QR}}{\text{PR}}$

$\sin P = \frac{12}{13}$

Let's find $\cos P$.

$\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

$\cos P = \frac{\text{PQ}}{\text{PR}}$

$\cos P = \frac{5}{13}$

Let's find $\tan P$.

$\tan P = \frac{\text{Opposite}}{\text{Adjacent}}$

$\tan P = \frac{\text{QR}}{\text{PQ}}$

$\tan P = \frac{12}{5}$

Answer

(i) $\sin P = \frac{12}{13}$ (ii) $\cos P = \frac{5}{13}$ (iii) $\tan P = \frac{12}{5}$