Question: Question 7 (3 points)\n\nSolve for $\angle A$. Round your answer to the nearest degrees.\n\nShow all your work!\n\n62cm\n51cm\n48cm\n\nA\n?\nB\nC\n\nTriangle ABC with sides $AB = 62$ cm, $AC = 51$ cm, and $BC = 48$ cm; angle $A$ is at the top vertex between sides $AB$ and $AC$. position_hint = center\n\nQuestion 8 (5 points)\n\nThe perimeter of a rectangle is $8$ m, and its area is $2$ m$^2$.\n\nFind the length and width of the rectangle to the nearest tenth of a metre.\n\n(Hint: Define your expressions, and then create an equation to solve.)\n\nLet $x$ be the length.\n\nLet __________ be the width.\n\n$x$\n\nA rectangle is shown in the lower-left area, with the top horizontal side labeled $x$.\n\nposition_hint: bottom-left\n\nQuestion 9 (5 points)\n\nFrom a certain point, the angle of elevation to the top of the church steeple is $32^\circ$.\n\nAt a point $125$ m closer to the steeple, the angle of elevation is $57^\circ$.\n\nCalculate the height of the church, to the nearest tenth of a metre.\n\nShow all your work!\n\nGraph/shape: A right-triangle style diagram of a church steeple, with two lines of sight from two points on the ground to the top of the steeple. The left ground point is farther away and marked with angle $32^\circ$ at bottom-left; a second ground point $125$ m to the right is marked with angle $57^\circ$; the steeple is on the right side of the diagram. position_hint: center
1. **Problem 7: Find $\angle A$ in triangle $ABC$ with sides $AB=62$ cm, $AC=51$ cm, and $BC=48$ cm.**\n\n2. Use the Law of Cosines formula to find angle $A$:\n$$\cos A = \frac{AB^2 + AC^2 - BC^2}{2 \times AB \times AC}$$\n\n3. Substitute the values:\n$$\cos A = \frac{62^2 + 51^2 - 48^2}{2 \times 62 \times 51} = \frac{3844 + 2601 - 2304}{6324} = \frac{4141}{6324}$$\n\n4. Calculate the fraction:\n$$\cos A \approx 0.6549$$\n\n5. Find $A$ by taking the inverse cosine:\n$$A = \cos^{-1}(0.6549) \approx 49.2^\circ$$\n\n6. Rounded to the nearest degree, $\angle A = 49^\circ$.\n\n---\n\n7. **Problem 8: Find length and width of a rectangle with perimeter $8$ m and area $2$ m$^2$.**\n\n8. Let $x$ be the length and $w$ be the width.\n\n9. The perimeter formula is:\n$$2(x + w) = 8 \implies x + w = 4$$\n\n10. The area formula is:\n$$x \times w = 2$$\n\n11. From perimeter, express $w$ in terms of $x$:\n$$w = 4 - x$$\n\n12. Substitute into area equation:\n$$x(4 - x) = 2$$\n$$4x - x^2 = 2$$\n\n13. Rearrange to quadratic form:\n$$x^2 - 4x + 2 = 0$$\n\n14. Use quadratic formula:\n$$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times 2}}{2} = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2}$$\n\n15. Simplify $\sqrt{8} = 2\sqrt{2} \approx 2.828$:\n$$x = \frac{4 \pm 2.828}{2}$$\n\n16. Calculate both solutions:\n$$x_1 = \frac{4 + 2.828}{2} = 3.414$$\n$$x_2 = \frac{4 - 2.828}{2} = 0.586$$\n\n17. Corresponding widths:\n$$w_1 = 4 - 3.414 = 0.586$$\n$$w_2 = 4 - 0.586 = 3.414$$\n\n18. So the length and width are approximately $3.4$ m and $0.6$ m (to nearest tenth).\n\n---\n\n19. **Problem 9: Calculate height of church steeple given two angles of elevation and distance difference.**\n\n20. Let $h$ be the height of the steeple, and $d$ be the distance from the farther point to the base of the steeple.\n\n21. From the farther point, angle of elevation is $32^\circ$:\n$$\tan 32^\circ = \frac{h}{d} \implies h = d \tan 32^\circ$$\n\n22. From the closer point (distance $d - 125$ m), angle of elevation is $57^\circ$:\n$$\tan 57^\circ = \frac{h}{d - 125} \implies h = (d - 125) \tan 57^\circ$$\n\n23. Set the two expressions for $h$ equal:\n$$d \tan 32^\circ = (d - 125) \tan 57^\circ$$\n\n24. Expand and rearrange:\n$$d \tan 32^\circ = d \tan 57^\circ - 125 \tan 57^\circ$$\n$$d \tan 32^\circ - d \tan 57^\circ = -125 \tan 57^\circ$$\n$$d (\tan 32^\circ - \tan 57^\circ) = -125 \tan 57^\circ$$\n\n25. Solve for $d$:\n$$d = \frac{-125 \tan 57^\circ}{\tan 32^\circ - \tan 57^\circ}$$\n\n26. Calculate values:\n$$\tan 32^\circ \approx 0.6249$$\n$$\tan 57^\circ \approx 1.5399$$\n\n27. Substitute:\n$$d = \frac{-125 \times 1.5399}{0.6249 - 1.5399} = \frac{-192.49}{-0.915} \approx 210.3$$\n\n28. Calculate height $h$:\n$$h = d \tan 32^\circ = 210.3 \times 0.6249 \approx 131.3$$\n\n29. Rounded to nearest tenth, the height of the church is $131.3$ m.