Problem about triangle

(1)By Pythagoras thm, triangle ABC is a right-angled triangle with angle ABC = 90.
and angle AOB = 90- x. Therefore, in triangle ABO, angle AOB = 180 - x -(90 -x ) = 90.
(2) angle ACB = arctan(3/4) = 36.9, therefore, angle OCB = 36.9 - x. Therefore, in triangle OBC, angle COB = 180 - x -(36.9 - x) = 143.1
(3) Angle AOC = 360 - 90 - 143.1 = 126.9
(4)Let OA = p, OB = r and OC = s.
Apply sine rule to triangle OBC, we get 4/sin143.1 = s/sinx................(a)
Apply sine rule to triangle AOC, we get 5/sin126.9 = p/sinx................(b)
Apply sine rule to triangle AOB, we get 3/sin90 = r/sinx......................(c)
(5) (c)/(b), (3/sin90)/(5/sin126.9) = (r/sinx)/(p/sinx) = r/p = tanx.
Therefore, tanx = 3sin126.9/5 = 0.48. And x = arctan(0.48) = 25.6.
(6)From (a), s = OC = 4sin25.6/sin143.1 = 2.88.
From (b), p = OA = 5sin25.6/sin126.9 = 2.70.
From (c), r = OB = 3sin25.6 = 1.30

2008-06-18 19:26:46 補充：
r/p = tanx because r/p = OB/OA and angle BOA = 90.