Question: 3
Answer all the questions.
P
9
Q
12
R
15
Triangle PQR is a right-angled triangle.
PQ = 9 cm, QR = 12 cm and PR = 15 cm.
Express as a fraction in the simplest form, the value of
(a) sin \angle PRQ,
sin \angle PRQ = 9/15
\angle PRQ = \sin^{-1}(9/15)
\angle PRQ = 36.870 (5sf)
\angle PRQ = 36.9^\circ (3sf)
Answer 36.9^\circ [1]
(b) tan \angle QPR.
tan \angle QPR = 12/9
\angle QPR = \tan^{-1}(12/9)
\angle QPR = 53.130 (5sf)
\angle QPR = 53.1^\circ (3sf)
Answer 53.1^\circ [1]
graph shape: right-angled triangle PQR with the right angle at Q, sides PQ = 9, QR = 12, and hypotenuse PR = 15; position_hint=center
1. **Problem Statement:** We have a right-angled triangle PQR with right angle at Q.
Given sides: $PQ = 9$, $QR = 12$, and hypotenuse $PR = 15$.
2. **Part (a): Find $\sin \angle PRQ$ as a fraction in simplest form and its angle measure.**
- The angle $\angle PRQ$ is opposite side $PQ$ and hypotenuse is $PR$.
- Formula: $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
- Substitute values: $\sin \angle PRQ = \frac{PQ}{PR} = \frac{9}{15}$
- Simplify fraction:
$$\frac{9}{15} = \frac{\cancel{3} \times 3}{\cancel{3} \times 5} = \frac{3}{5}$$
- Calculate angle:
$$\angle PRQ = \sin^{-1}\left(\frac{3}{5}\right) \approx 36.870^\circ$$
- Rounded to 3 significant figures:
$$36.9^\circ$$
3. **Part (b): Find $\tan \angle QPR$ and its angle measure.**
- The angle $\angle QPR$ is opposite side $QR$ and adjacent side $PQ$.
- Formula: $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
- Substitute values: $\tan \angle QPR = \frac{QR}{PQ} = \frac{12}{9}$
- Simplify fraction:
$$\frac{12}{9} = \frac{\cancel{3} \times 4}{\cancel{3} \times 3} = \frac{4}{3}$$
- Calculate angle:
$$\angle QPR = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.130^\circ$$
- Rounded to 3 significant figures:
$$53.1^\circ$$
**Final answers:**
- (a) $\sin \angle PRQ = \frac{3}{5}$ and $\angle PRQ = 36.9^\circ$
- (b) $\tan \angle QPR = \frac{4}{3}$ and $\angle QPR = 53.1^\circ$