Question

Given 5tan2A = 2tanA + 6, find A to the nearest tenth of a degree when 180º≤A≤270º.

Answer 1

5(tanA)^2 = 2tanA + 6
5(tanA)^2 - 2tanA - 6 =0
apply quadratic formula (-b+- sqrt(b^2-4ac))/2a, we have
(-(-2)+- sqrt((-2)^2 - 4(5)(-6)))/2(5) = tan A
tan A = (1+- sqrt31)/5
so you compute A = tan^-1 (1+- sqrt31)/5 and get the answer
you will see 52.72 from your calculator, but since A is betwen 180 and 270, so you should add 180, which gets 232.72 = 232.7

Answer 2

OMG thank you ! can you solve this Find tan2A when cosA = -3/5 and π/2<A<π.