1. **Problem statement:** - Question 6: In triangle KLM, KM = 6 cm and angle KLM = 30°. Find the length of KL. - Question 7: In triangle PQR, angle QPR = 45° and PQ = 12 cm. Find the length of PR. 2. **Relevant formula:** For right triangles, use the trigonometric ratios: - $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$ - $$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$$ - $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$ 3. **Question 6 solution:** - Given: Right triangle KLM with right angle at L (implied by the problem), KM = 6 cm, angle KLM = 30°. - We want to find KL. - In triangle KLM, angle KLM is at vertex L, so KL is adjacent to angle 30°, and KM is the hypotenuse. - Using cosine: $$\cos 30^\circ = \frac{KL}{KM}$$ - Substitute known values: $$\cos 30^\circ = \frac{KL}{6}$$ - Multiply both sides by 6: $$KL = 6 \times \cos 30^\circ$$ - Recall $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$ - So: $$KL = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}$$ - Simplify: $$KL = 3\sqrt{3} \text{ cm}$$ 4. **Question 7 solution:** - Given: Right triangle PQR with right angle at Q, angle QPR = 45°, and PQ = 12 cm. - We want to find PR. - Angle QPR is at vertex P, so PQ is adjacent to angle 45°, and PR is the hypotenuse. - Using cosine: $$\cos 45^\circ = \frac{PQ}{PR}$$ - Substitute known values: $$\cos 45^\circ = \frac{12}{PR}$$ - Multiply both sides by PR: $$PR \times \cos 45^\circ = 12$$ - Divide both sides by $$\cos 45^\circ$$: $$PR = \frac{12}{\cos 45^\circ}$$ - Recall $$\cos 45^\circ = \frac{\sqrt{2}}{2}$$ - So: $$PR = \frac{12}{\frac{\sqrt{2}}{2}} = 12 \times \frac{2}{\sqrt{2}} = \frac{24}{\sqrt{2}}$$ - Rationalize denominator: $$PR = \frac{24}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{24\sqrt{2}}{2} = 12\sqrt{2}$$ - So: $$PR = 12\sqrt{2} \text{ cm}$$ **Final answers:** - Question 6: KL = $3\sqrt{3}$ cm - Question 7: PR = $12\sqrt{2}$ cm