Posted by suz on Tuesday, January 11, 2011 at 3:50pm.
I worked it backwards, since the second equation was easier to solve
tan^2Ø + tanØ-6=0
(tanØ-2)(tanØ+3) = 0
tanØ = 2 or tanØ=-3
case1: tanØ=2, Ø could be in I or III
In I, sinØ = 2/√5 and cosØ = 1/√5
test in first equation
LS = 7(4/5) + (1/√5)(2/√5)
= 28/5 + 2/5 = 30/5 = 6 = RS
In III
sinØ = -2/√5 , cosØ = -1/√5
LS = 7(4/5) + (-2/√5)(-1/√5) = 30/5 = 6 = RS
Case 2: tanØ = -3, Ø could be in II or IV
in II , sinØ = 3/√10 , cosØ = -1/√10
testing in first equation,
LS = 7(9/10) + (-1/√10)((3/√10)
= 63/10 - 3/10 = 60/10 = 6 = RS
in IV, sinØ = -3/√10 , cosØ = 1/√10
LS = 7(9/10) + (1/√10)(-3/√10)
= 63/10 - 3/10 = 60/10 = 6 = RS
so all 4 values of Ø satisfy both equations.