Question

How do you find θ, if `0 < θ < 360` and `tan theta = sqrt3` and theta in in QIII?

Answer 1

In QIII `sin theta < 0` and `cos theta < 0`.

Notice that:

`sqrt(3) = (-sqrt(3)/2) / (-1/2)`

`= (-sin 60)/(-cos 60)`

`= sin (180+60)/cos(180+60)`

`= (sin 240) / (cos 240)`

`=tan 240`

To see that `cos 60 = 1/2`, picture an equilateral triangle with sides of length 1 and cut it in half to produce two right angled triangles with internal angles 30, 60 and 90 degrees. The length of the shortest side is 1/2, the length of the hypotenuse is 1 and the length of the other side will be:

`sqrt(1^2 - (1/2)^2) = sqrt(1-1/4) = sqrt(3/4) = sqrt(3)/2`