Trigonometry
posted by John .
How do you solve the equation
5√3 secθ = 10 in degree measure?
please explain.

secØ = 10/5√3 = 2/√3
then
cosØ = √3/2
You should be familiar with the 1 : √3 : 2 rightangled triangle which has angles of 30:60:90
thus Ø = 30°
but the cosine is also positive in the fourth quadrant
so Ø could be 330°
Ø = 30° or Ø = 330°