Asked by Anonymous
Find all angles, 0≤A<360, that satisfy the equation below, to the nearest 10th of a degree.
2tanA+6=tanA+3
2tanA+6=tanA+3
Answers
Answered by
Reiny
2tanA+6=tanA+3
tanA = -3
so A must be in II or IV
using tanx = +3
x = 71.565°
so A = 180-71.564 = appr 108.4
or
A = 360 - 71.565 = appr 288.4°
tanA = -3
so A must be in II or IV
using tanx = +3
x = 71.565°
so A = 180-71.564 = appr 108.4
or
A = 360 - 71.565 = appr 288.4°