1. **Problem (a):** Given a right-angled triangle with sides 3 (vertical), 4 (base), and 5 (hypotenuse), and angle $\beta$ at the bottom-right vertex, find the sum $\sin \beta + \tan \beta$ as a fraction. 2. **Recall the definitions:** - $\sin \beta = \frac{\text{opposite}}{\text{hypotenuse}}$ - $\tan \beta = \frac{\text{opposite}}{\text{adjacent}}$ 3. **Identify sides relative to $\beta$:** - Opposite side to $\beta$ is vertical side = 3 - Adjacent side to $\beta$ is base = 4 - Hypotenuse = 5 4. **Calculate each trigonometric ratio:** $$\sin \beta = \frac{3}{5}$$ $$\tan \beta = \frac{3}{4}$$ 5. **Sum the two:** $$\sin \beta + \tan \beta = \frac{3}{5} + \frac{3}{4}$$ 6. **Find common denominator and add:** $$\frac{3}{5} + \frac{3}{4} = \frac{3 \times 4}{5 \times 4} + \frac{3 \times 5}{4 \times 5} = \frac{12}{20} + \frac{15}{20} = \frac{27}{20}$$ --- 7. **Problem (b)(i):** Triangle $PQR$ with $|PQ|=5$ cm, $|QR|=9$ cm, and area $=13.71$ cm$^2$. Find angle $\angle PQR$ to the nearest degree. 8. **Recall area formula for triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 9. **Here, base $QR=9$ cm. Let height from $P$ to $QR$ be $h$. Then:** $$13.71 = \frac{1}{2} \times 9 \times h \implies h = \frac{2 \times 13.71}{9} = \frac{27.42}{9} = 3.0467$$ 10. **Use right triangle formed by height $h$ and side $PQ=5$ cm to find angle $\angle PQR$:** - $\sin \angle PQR = \frac{h}{PQ} = \frac{3.0467}{5} = 0.60934$ 11. **Calculate angle:** $$\angle PQR = \arcsin(0.60934) \approx 37.5^\circ$$ 12. **Rounded to nearest degree:** $$\angle PQR \approx 38^\circ$$ --- 13. **Problem (b)(ii):** Find length $|PR|$ to nearest cm. 14. **Use Law of Cosines:** $$|PR|^2 = |PQ|^2 + |QR|^2 - 2 \times |PQ| \times |QR| \times \cos \angle PQR$$ 15. **Calculate $\cos \angle PQR$:** $$\cos 38^\circ \approx 0.7880$$ 16. **Calculate $|PR|^2$:** $$|PR|^2 = 5^2 + 9^2 - 2 \times 5 \times 9 \times 0.7880 = 25 + 81 - 90 \times 0.7880 = 106 - 70.92 = 35.08$$ 17. **Calculate $|PR|$:** $$|PR| = \sqrt{35.08} \approx 5.92$$ 18. **Rounded to nearest cm:** $$|PR| \approx 6 \text{ cm}$$