1. Problem 4 statement: In right triangle ABC the right angle is at C, the hypotenuse AB = $2$ cm, and angle at B = $45^\circ$. 2. Since the triangle is right angled at C and angle B = $45^\circ$, angle A = $90^\circ - 45^\circ = 45^\circ$. 3. Therefore the triangle is a $45^\circ$-$45^\circ$-$90^\circ$ isosceles right triangle, so the two legs are equal. 4. For a $45^\circ$-$45^\circ$-$90^\circ$ triangle the leg length equals the hypotenuse divided by $\sqrt{2}$, so each leg is $\dfrac{2}{\sqrt{2}}$ cm. 5. Simplify $\dfrac{2}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$ cm. 6. Therefore $a = \sqrt{2}$ cm and $b = \sqrt{2}$ cm. 7. Problem 5 statement: In right triangle PQR the right angle is at Q, PQ = $4$, hypotenuse PR = $q$, and angle at R = $45^\circ$. 8. Since angle R = $45^\circ$ the triangle is $45^\circ$-$45^\circ$-$90^\circ$, so hypotenuse = $\text{leg}\cdot\sqrt{2}$. 9. Thus $q = 4\sqrt{2}$. 10. Therefore $q = 4\sqrt{2}$. 11. Problem 6 statement: In isosceles triangle XYZ we have XY = XZ = $6$ cm and angle YXZ = $120^\circ$, find YZ. 12. Use the law of cosines: $YZ^2 = XY^2 + XZ^2 - 2\,XY\,XZ\cos(\angle YXZ)$. 13. Substitute values: $YZ^2 = 6^2 + 6^2 - 2\cdot 6 \cdot 6 \cos(120^\circ)$. 14. Evaluate $\cos(120^\circ) = -\tfrac{1}{2}$. 15. So $YZ^2 = 36 + 36 - 72 \cdot (-\tfrac{1}{2}) = 72 + 36 = 108$. 16. Therefore $YZ = \sqrt{108} = 6\sqrt{3}$ cm. 17. Problem 7 statement: A cone has perpendicular height $42$ cm and semi-vertical angle $30^\circ$, find the slant height. 18. If the semi-vertical angle is the angle between the slant height and the axis, then $\cos(30^\circ) = \dfrac{\text{height}}{\text{slant height}} = \dfrac{42}{l}$. 19. Solve for slant height $l = \dfrac{42}{\cos(30^\circ)}$. 20. Since $\cos(30^\circ) = \tfrac{\sqrt{3}}{2}$ we get $l = \dfrac{42}{\tfrac{\sqrt{3}}{2}} = \dfrac{42 \cdot 2}{\sqrt{3}} = \dfrac{84}{\sqrt{3}}$. 21. Rationalize if desired: $l = \dfrac{84}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{84\sqrt{3}}{3} = 28\sqrt{3}$ cm. 22. Therefore the slant height is $28\sqrt{3}$ cm.