a.maths trigonometry

Note: I have used x instead of theta to represent the angle for easy typing.
Q.1
Angle EBD = angle BDC - angle BED = 2x - (90 - x ) = (3x - 90). (Exterior angle of triangle.)
Q.2
Let AB = m.
For triangle BAE, applying sine rule, we get
m/sin[180 -(90-x)] = a/sin(angle ABE). Angle ABE = 90 - x - x = (90 -2x) (Exterior angle of triangle).
Therefore, m/sin(90+x) = a/sin(90 -2x). That is m/cosx = a/cos2x.........(1)
For triangle ABD, applying sine rule, we get
m/sin(180-2x) = (a+b)/sin(angleABD). Angle ABD = 2x -x = x (Exterior angle of triangle).
That is m/sin(180 -2x) = (a+b)/sinx
Or m/sin2x = (a+b)/sinx..................................(2)
(1) /(2) we get sin2x/cosx = asinx/[(a+b)cos2x]
2sinx = asinx/[(a+b)cos2x]
Therefore, cos2x = a/[2(a+b)].

問題的中心係幾個基礎三角形定律
上半題:
(三角形內角和180度)
角EBD = 180 - ( 90 -X) - (180 -2X) = 3X - 90
下半題:
(三角形內角和180度)
(sin rule)
同樣地,角ABE = 180 -X -[180 -(90-X)] = 90 + 2X
SO, 角ABD = 3X -90 +90 -2X = X
SO, 三角形ABD 為等腰三角形
AD = DB = a+b

sin rule:
sin角ABE/a = sin角BAE/BE
sin(90-2X)/a = sinX/BE
cos2x = a*sinX /BE

sin角BED/BD = sin角BDE/BE
sin(90 - X)/(a+b) = sin (180 -2X)/BE
BE = sin2X*(a+b)/cosX
BE = 2sinXcosX*(a+b)/cosX
BE = 2sinX*(a+b)

so, cos2X =[a*sinX]/[2sinX*(a+b)]
cos2X = a/2(a+b)