Question: FINAL EXAM Page 1: 1 -- Page 2: 2 -- Page 3: 3 -- Page 4: 4 -- Page 5: 5 -- Page 6: Question 6 (4 points) a) Determine the value of $x$. $x$ $48m$ $18^\circ$ Graph shape: a right triangle with the right angle at the top-left vertex, the bottom side labeled $48m$, the left side labeled $x$, and the angle at the right tip labeled $18^\circ$. position_hint=top-center b) Determine the measure of $\angle A$ to the nearest degree. $A$ $B$ $C$ $3m$ $7m$ Graph shape: a right triangle with a right angle at vertex $B$, segment $AB$ labeled $3m$ across the top, segment $AC$ labeled $7m$ on the left side, and point $C$ at the bottom. position_hint=center Show all your work!Question 4 (6 points) Determine an equation of the graph below in the following forms: a) in factored form, $y = a(x - r)(x - s)$ b) in vertex form, $y = a(x - h)^2 + k$ Show all your work! Graph shape: downward-opening parabola; x-intercepts at $(-4, 0)$ and $(8, 0)$; vertex at $(2, 6)$. position_hint: center Question 9 (5 points) From a certain point, the angle of elevation to the top of the church steeple is $32^\circ$. At a point $125m$ closer to the steeple, the angle of elevation is $57^\circ$. Calculate the height of the church, to the nearest tenth of a metre. Show all your work! Graph: a right-triangle elevation diagram in the bottom-center of the page, with a church steeple on the right, two sight lines from two ground points to the top, the left angle labeled $32^\circ$ and the nearer angle labeled $57^\circ$, and the distance between the two ground points labeled $125m$; position_hint \in {bottom-left, bottom-right, center} with the diagram centered toward the bottom.

### Problem 6a: Determine the value of $x$ in the right triangle 1. **Problem statement:** Given a right triangle with a right angle at the top-left vertex, the bottom side length $48m$, the left side length $x$, and an angle of $18^\circ$ at the right tip, find $x$. 2. **Formula used:** In a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ 3. **Identify sides:** For the $18^\circ$ angle, the side opposite is $x$, and the adjacent side is $48m$. 4. **Set up the equation:** $$\tan(18^\circ) = \frac{x}{48}$$ 5. **Solve for $x$:** $$x = 48 \times \tan(18^\circ)$$ 6. **Calculate:** $$x = 48 \times 0.3249 = 15.5952$$ 7. **Final answer:** $$x \approx 15.6m$$ --- ### Problem 6b: Determine the measure of $\angle A$ in the right triangle 1. **Problem statement:** Given a right triangle with right angle at $B$, side $AB = 3m$, side $AC = 7m$, find the measure of $\angle A$. 2. **Formula used:** Use the tangent function: $$\tan(\angle A) = \frac{\text{opposite}}{\text{adjacent}}$$ 3. **Identify sides:** Opposite side to $\angle A$ is $AB = 3m$, adjacent side is $BC$ which we need to find first. 4. **Find $BC$ using Pythagoras:** $$BC = \sqrt{AC^2 - AB^2} = \sqrt{7^2 - 3^2} = \sqrt{49 - 9} = \sqrt{40} = 6.3246$$ 5. **Calculate tangent:** $$\tan(\angle A) = \frac{3}{6.3246} = 0.4743$$ 6. **Find angle:** $$\angle A = \tan^{-1}(0.4743) = 25.4^\circ$$ 7. **Final answer:** $$\angle A \approx 25^\circ$$ --- ### Problem 4: Determine the equation of the parabola 1. **Problem statement:** Find the equation of a downward-opening parabola with x-intercepts at $-4$ and $8$, and vertex at $(2,6)$. 2. **Factored form:** $$y = a(x - r)(x - s)$$ where $r = -4$, $s = 8$. 3. **Substitute intercepts:** $$y = a(x + 4)(x - 8)$$ 4. **Use vertex to find $a$:** At vertex $(2,6)$: $$6 = a(2 + 4)(2 - 8) = a(6)(-6) = -36a$$ 5. **Solve for $a$:** $$a = \frac{6}{-36} = -\frac{1}{6}$$ 6. **Factored form equation:** $$y = -\frac{1}{6}(x + 4)(x - 8)$$ 7. **Vertex form:** $$y = a(x - h)^2 + k$$ where vertex $(h,k) = (2,6)$ and $a = -\frac{1}{6}$. 8. **Vertex form equation:** $$y = -\frac{1}{6}(x - 2)^2 + 6$$ --- ### Problem 9: Calculate the height of the church steeple 1. **Problem statement:** From two points on the ground, the angles of elevation to the top of the church steeple are $32^\circ$ and $57^\circ$. The points are $125m$ apart. Find the height $h$ of the steeple. 2. **Label points:** Let the farther point be $P$, the nearer point $Q$, and the steeple base $S$. Distance $PQ = 125m$. 3. **Set variables:** Let $h$ be the height of the steeple, and $x$ be the horizontal distance from $Q$ to $S$. 4. **From $Q$ (nearer):** $$\tan(57^\circ) = \frac{h}{x} \Rightarrow h = x \tan(57^\circ)$$ 5. **From $P$ (farther):** Distance from $P$ to $S$ is $x + 125$. $$\tan(32^\circ) = \frac{h}{x + 125} \Rightarrow h = (x + 125) \tan(32^\circ)$$ 6. **Set equal:** $$x \tan(57^\circ) = (x + 125) \tan(32^\circ)$$ 7. **Expand:** $$x \tan(57^\circ) = x \tan(32^\circ) + 125 \tan(32^\circ)$$ 8. **Group $x$ terms:** $$x (\tan(57^\circ) - \tan(32^\circ)) = 125 \tan(32^\circ)$$ 9. **Solve for $x$:** $$x = \frac{125 \tan(32^\circ)}{\tan(57^\circ) - \tan(32^\circ)}$$ Calculate values: $$\tan(32^\circ) \approx 0.6249$$ $$\tan(57^\circ) \approx 1.5399$$ $$x = \frac{125 \times 0.6249}{1.5399 - 0.6249} = \frac{78.1125}{0.915} = 85.36m$$ 10. **Calculate height $h$:** $$h = x \tan(57^\circ) = 85.36 \times 1.5399 = 131.4m$$ 11. **Final answer:** $$h \approx 131.4m$$