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