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

1 Answer
George C.
May 12, 2015

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#

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