F.2幾何(long questions)(20分)

Q1. Method 1 : Since height of the 2 triangles are the same ( which is the perpendicular distance from C to AB), so if their base is the same, area will be the same, that is AD = DB. By Pythagoras theorem AB = 13, so AD = DB = 13/2 = 6.5
Method 2 : Let angle CAD = x, so tan x = 12/5 ( that is sin x = 12/13 and cos x = 5/13).
Height of triangle = 5 sin x = 5(12/13) = 60/13.
Area of triangle ABC = (5)(12)/2 = 30.
so (60/13)(AD)/2 = half of 30
so 30AD/13 = 15
AD = 13 x 15/30 = 6.5
DB = AB - AD = 13 - 6.5 = 6.5
angle CAB = arc tan (12/5)
angle CBA = 180 - 90 - angle CAB.
Q2.
Let CD = h, h/AD = tan t ( t = theta).
h/BD = tan y ( y = phi)
so tan y/tan t = (h/BD)/(h/AD) = AD/BD = AD/3AD = 1/3
so tan y = (tan t)/3................[answer for (b)]
cos x = AE/AB = sin (y + x).........[answer for (a)]
(ci) If y = 21, from (b) tan 21 = (tan t)/3
so t = arc tan (3tan 21)




2009-04-10 23:08:08 補充：
Q2. AC = AD/cos t, BC = BD/cos y = 3AD/cos y, so AD(1/cos t + 3/cos y + 4) = 17, from this equation we can get the value of AD, so AC = AD/cos t. x = 90 - t - y.