2 Maths CE Qs

2003(a) cos∠OAC = (3^2 + 6^2 - 4^2)/(2 * 3 * 6)
∠OAC = 36.3360575(b)(i) BC/OC = tan40,BC = OC tan40 = 3.3562 m
BC/CD = tan30
CD = BC/tan30
CD = 3.3562/tan30
CD = 5.8131 m(ii) cos∠CAD = (6^2 + 8^2 - 5.8131^2)/(2 * 6 * 8)
∠CAD = 46.39638(iii) cos∠ACD = (6^2 + 5.8131^2 - 8^2)/(2 * 6 * 5.8131)
∠CAD = 85.2371So, θ = 180 - ∠OAC - ∠CAD = 58.42682005(a) BE/BC = sin30
BE = BC sin30 = 60 cmCE/BC = cos30
CE = BC cos30 = 103.923048 cm(b) ∠ABC = 80, ∠CAB = 60, ∠ACB = 40AB/sin∠ACB = BC/sin∠CAB
AB/sin40 = 120/sin60
AB = 89.0673 cmAC/sin∠ABC = BC/sin∠CAB
AC/sin80 = 120/sin60
AC = 136.459 cm(c) sin∠ACD = AD/AC = 100/136.459
∠ACD = 47.1234AD/CD = tan∠ACD
CD = AD/tan∠ACD = 92.8497 cmDraw a line from B to AD. Call the intersection point F
Then AF = 40
BF^2 = AB^2 - AF^2 => BF = 79.58 cm
So, DE = 79.58 cmcos∠CDE = (DE^2 + CD^2 - CE^2)/(2 * DE * CD)
cos∠CDE = (79.58^2 + 92.8497^2 - 103.923048^2)/(2 * 79.58 * 92.8497)
∠CDE = 73.6743The shortest distance should be CG such that CG is prependicular to DE
As ∠CDE = 73.6743
CG/CD = sin∠CDE
CG = CD sin∠CDE
CG = 92.8497 sin 73.6743
CG = 89.1059 cm