1. **Problem 1: Find the length of $x$ in two similar triangles.** Given triangles $\triangle JKL$ and $\triangle MNL$ are similar. Sides in smaller triangle: $JK=9$ mm, $JL=13$ mm. Sides in larger triangle: $MN=27$ mm, $ML=x$ mm. Since triangles are similar, corresponding sides are proportional: $$\frac{JK}{MN} = \frac{JL}{ML}$$ Substitute known values: $$\frac{9}{27} = \frac{13}{x}$$ 2. Cross multiply: $$9 \times x = 27 \times 13$$ $$9x = 351$$ 3. Divide both sides by 9: $$x = \frac{351}{9}$$ Show cancellation: $$x = \frac{\cancel{351}}{\cancel{9}} = 39$$ So, $x = 39$ mm. --- 2. **Problem 2: Solve for unknown variables using SOHCAHTOA.** (a) Find $x$ to one decimal place in right triangle with angle $35^\circ$, opposite side $x$, adjacent side $15$ m. Use tangent since opposite and adjacent sides: $$\tan(35^\circ) = \frac{x}{15}$$ Multiply both sides by 15: $$x = 15 \times \tan(35^\circ)$$ Calculate: $$x \approx 15 \times 0.7002 = 10.5$$ So, $x \approx 10.5$ m. (b) Find $b$ to one decimal place in right triangle with angle $40^\circ$, adjacent side $12$ cm, hypotenuse $b$. Use cosine since adjacent and hypotenuse: $$\cos(40^\circ) = \frac{12}{b}$$ Cross multiply: $$b \times \cos(40^\circ) = 12$$ Divide both sides by $\cos(40^\circ)$: $$b = \frac{12}{\cos(40^\circ)}$$ Show cancellation: $$b = \frac{12}{\cancel{\cos(40^\circ)}} = 12 \div 0.7660$$ Calculate: $$b \approx 15.7$$ So, $b \approx 15.7$ cm. --- 3. **Problem 3: Determine unknown side length using Pythagorean theorem.** (a) Right triangle with legs 6 and 8, hypotenuse $x$. Pythagorean theorem: $$x^2 = 6^2 + 8^2 = 36 + 64 = 100$$ Take square root: $$x = \sqrt{100} = 10$$ (b) Right triangle with legs 3 and $x$, hypotenuse 12. Pythagorean theorem: $$12^2 = 3^2 + x^2$$ $$144 = 9 + x^2$$ Subtract 9: $$x^2 = 144 - 9 = 135$$ Take square root: $$x = \sqrt{135} = \sqrt{9 \times 15} = 3\sqrt{15} \approx 11.6$$ --- **Final answers:** 1) $x = 39$ mm 2a) $x \approx 10.5$ m 2b) $b \approx 15.7$ cm 3a) $x = 10$ 3b) $x \approx 11.6$