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November 23, 2014

November 23, 2014

Posted by **suz** on Tuesday, January 11, 2011 at 3:50pm.

- calculus -
**Reiny**, Tuesday, January 11, 2011 at 5:01pmI 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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