1. **Problem statement:** Calculate various geometric and trigonometric properties of the tent roof and complete a table for the cubic function. ### Question 19 (a) Find the length of the projection of FC on the ground. - Given: FC = 6 m, height of post P3 = 2.5 m. - The projection of FC on the ground is the horizontal component of FC. - Using Pythagoras theorem in triangle formed by FC, its vertical height, and its projection: $$\text{Projection of } FC = \sqrt{FC^2 - (\text{height})^2} = \sqrt{6^2 - 2.5^2} = \sqrt{36 - 6.25} = \sqrt{29.75} \approx 5.455 \text{ m}$$ (b) Height of ridge EF above the ground. - EF = 3 m is centrally placed on the roof. - Posts are 2.5 m high. - Since EF is the ridge, its height is the post height plus the vertical height of the roof above the posts. - Using triangle with edges FB = 6 m and horizontal distance FB projection (from part a), vertical height of roof above post is: $$\sqrt{6^2 - 5.455^2} = \sqrt{36 - 29.75} = \sqrt{6.25} = 2.5 \text{ m}$$ - Total height of ridge EF above ground = post height + roof height = $2.5 + 2.5 = 5.0$ m. (c) Angle between edges FB and FC. - Both FB and FC are 6 m. - The angle between FB and FC can be found using the cosine rule in triangle FBC. - Given BC = 8 m. $$\cos \theta = \frac{FB^2 + FC^2 - BC^2}{2 \times FB \times FC} = \frac{6^2 + 6^2 - 8^2}{2 \times 6 \times 6} = \frac{36 + 36 - 64}{72} = \frac{8}{72} = \frac{1}{9}$$ $$\theta = \cos^{-1}\left(\frac{1}{9}\right) \approx 83.62^\circ$$ (d) Angle between plane FBC and the ground. - The plane FBC contains edges FB, BC, and FC. - The angle between the plane and the ground is the angle between the normal to the plane and the vertical. - Using the height of posts and horizontal distances, the angle can be approximated by the angle between edge FB and its projection. - From (a), projection of FC is 5.455 m, height is 2.5 m. $$\tan \alpha = \frac{2.5}{5.455} \Rightarrow \alpha = \tan^{-1}\left(\frac{2.5}{5.455}\right) \approx 24.44^\circ$$ ### Question 20 (a) Complete the table for $y = 3x^3 + x^2 - 7x$ for missing values. Calculate missing y values: - For $x = -1.5$: $$y = 3(-1.5)^3 + (-1.5)^2 - 7(-1.5) = 3(-3.375) + 2.25 + 10.5 = -10.125 + 2.25 + 10.5 = 2.625 \approx 2.6$$ - For $x = -1$: $$y = 3(-1)^3 + (-1)^2 - 7(-1) = 3(-1) + 1 + 7 = -3 + 1 + 7 = 5$$ - For $x = -0.5$: $$y = 3(-0.5)^3 + (-0.5)^2 - 7(-0.5) = 3(-0.125) + 0.25 + 3.5 = -0.375 + 0.25 + 3.5 = 3.375 \approx 3.4$$ - For $x = 0.5$: $$y = 3(0.5)^3 + (0.5)^2 - 7(0.5) = 3(0.125) + 0.25 - 3.5 = 0.375 + 0.25 - 3.5 = -2.875 \approx -2.9$$ - For $x = 1.5$: $$y = 3(1.5)^3 + (1.5)^2 - 7(1.5) = 3(3.375) + 2.25 - 10.5 = 10.125 + 2.25 - 10.5 = 1.875 \approx 1.9$$ (b) Graphing instructions provided but no graph requested. ### Summary - (19a) Projection of FC on ground = $5.455$ m - (19b) Height of ridge EF above ground = $5.0$ m - (19c) Angle between FB and FC = $83.62^\circ$ - (19d) Angle between plane FBC and ground = $24.44^\circ$ - (20a) Completed y values: -1.5: 2.6, -1: 5, -0.5: 3.4, 0.5: -2.9, 1.5: 1.9