1. **Problem:** Work out the length of the missing side of the triangles given. 2. **Formula:** Use the Pythagorean theorem for right triangles: $$A^2 + B^2 = C^2$$ where $C$ is the hypotenuse. 3. **Triangle a:** Sides 3 cm and 4 cm, hypotenuse $a$. Calculate: $$a = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm}$$ 4. **Triangle b:** Isosceles right triangle with two sides 6 cm each, hypotenuse $c$. Calculate: $$c = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \approx 8.5 \text{ cm}$$ 5. **Triangle d:** Right triangle with sides 10 mm and 15 mm, missing side $d$ (hypotenuse). Calculate: $$d = \sqrt{10^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325} = 5\sqrt{13} \approx 18.0 \text{ mm}$$ 6. **Work out the value of $y$ in right triangles:** - Triangle b: angle 56°, side adjacent 14 cm, side opposite $y$ cm. Use tangent: $$\tan 56^\circ = \frac{y}{14} \Rightarrow y = 14 \times \tan 56^\circ$$ Calculate: $$y = 14 \times 1.4826 = 20.8 \text{ cm}$$ - Triangle c: angle 51°, side adjacent 18 cm, side opposite $y$ cm. Use tangent: $$\tan 51^\circ = \frac{y}{18} \Rightarrow y = 18 \times \tan 51^\circ$$ Calculate: $$y = 18 \times 1.2349 = 22.2 \text{ cm}$$ 7. **Calculate fluency values rounded to 1 decimal place:** - $$\cos 56^\circ = 0.559$$ - $$\sin^{-1} \frac{3}{4} = \sin^{-1} 0.75 = 48.6^\circ$$ - $$\tan^{-1} 0.3 = 16.7^\circ$$ **Final answers:** - Triangle a: $a = 5$ cm - Triangle b: $c \approx 8.5$ cm - Triangle d: $d \approx 18.0$ mm - Triangle b (value of $y$): $y \approx 20.8$ cm - Triangle c (value of $y$): $y \approx 22.2$ cm - $\cos 56^\circ = 0.6$ (rounded) - $\sin^{-1} 3/4 = 48.6^\circ$ - $\tan^{-1} 0.3 = 16.7^\circ$