F5 Maths 三角學 Trigonometry

By Heron's formula, area of triangle ACD
= sqrt [s(s - 12)(s - 12)(s - 15)] where s = (12 + 12 + 15)/2 = 19.5
= sqrt [ (19.5)(7.5)(7.5)(4.5)] = 70.2562
So (AD x CF)/2 = 70.2562
CF = (70.2562)(2)/(12) = 11.7094
AF = sqrt [ AC^2 - CF^2) ] = sqrt (144 - 11.7094^2) = 2.625
By cosine rule,
BD^2 = AB^2 + AD^2 - 2(AB)(AD) cos (angle BAD)
20^2 = 14^2 + 12^2 - 2(12)(14) cos ( angle BAD)
so cos ( angle BAD) = - 0.17857
Again by cosine rule,
BF^2 = AB^2 + AF^2 - 2(AB)(AF) cos ( angle BAD)
= 14^2 + 2.625^2 - 2(14)(2.625)(- 0.17857) = 216.0155
so BF = 14.6945
Again by cosine rule,
BC^2 = BF^2 + CF^2 - 2(BF)(CF) cos (angle BFC)
20^2 = 14.6945^2 + 11.7094^2 - 2(14.6945)(11.7094) cos ( angle BFC)
so cos (angle BFC) = - 0.13646
so angle BFC = 97.843 = 98 degree ( nearest degree)

2015-01-01 10:52:54 補充：
b, Angle BFC is not the angle between plane ACD and ABD because : (1) BF > BA, so BF is certainly not the shortest distance from B to line AD.

2015-01-01 10:57:01 補充：
(2) cos ( angle BAD) = - 0.17857, so angle BAD = 100.30 degree which is an obtuse angle. So the perpendicular distance from B to line AD cannot be inside the triangle.

2015-01-01 11:01:19 補充：
(3) By using the Heron's formula, you can calculate the area of triangle BAD, so you can find the perpendicular distance from B to line AD, if it is shorter than BF, then angle BFC will not be the angle between the 2 planes.