急! F5 Trigonometry 11q7

Q20.
(a) Area of triangle FBC = (1/2)(8)(8)sin 45 = 22.6274
The perpendicular distance from F to BC x BC x 1/2 = Area of triangle FBC
So the perpendicular distance from F to BC = 22.6274 x 2/8 = 5.6569
AF^2 = 7^2 + 8^2, so AF = 10.6301
So angle between FA and plane ABCD = arc sin ( 5.6569/10.6301) = 32.15 degree.
(b)
By Cosine Rule, FC^2 = 8^2 + 8^2 - 2(8)(8)cos 45.
So FC = 6.1229
AF = AC = 10.6301
By Cosine Rule
FC^2 = AF^2 + AC^2 - 2(AF)(AC)cos (angle FAC)
37.4903 = 113 + 113 - 2(113) cos ( angle FAC)
angle FAC = angle between line FA and line AC
= arc cos (188.5097/226)
= 33.4763 degree.
Q21.
(ai) Angle between line AB and plane = arc sin ( 5/13) = 22.6199 degree.
(aii) By Cosine Rule,
AC^2 = AB^2 + BC^2 - 2(AB)(BC) cos ( angle ABC)
8^2 = 8^2 + 13^2 - 2(13)(8) cos ( angle ABC)
so angle ABC = arc cos ( 13/16) = 35.6591 degree
so the perpendicular distance from A to BC = AB sin 35.6591 = 13 sin 35.6591 = 7.5785
so angle between the 2 planes = arc sin ( 5/7.5785) = 41.2817 degree.
(bi)
The perpendicular distance from D to BC/The perpendicular distance from A to BC = cos 41.2817
So the perpendicular distance from D to BC = 7.5785 x cos 41.2817 = 5.6951
So area of the projection = 5.6951 x 8 x 1/2 = 22.7802
(bii)
Since area of projection = area of triangle ABC x cos ( angle between ABC and the projection), so when A is lifted up, angle between them increases, cosine of the angle decreases, so area of projection decreases.