Alternatively, you could think of this as #tan(60˚)#, and then draw a #30˚-60˚-90˚# triangle:
#tan(60˚)# will be equal to #"opposite"/"adjacent"# in reference to the #60˚# angle, so we see that #"opposite"=sqrt3# and #"adjacent"=1#. Hence,
#tan(60˚)="opposite"/"adjacent"=sqrt3/1=sqrt3#
We can also examine the unit circle at #pi/3#:
If we know the point #(1/2,sqrt3/2)#, we can determine tangent if we think about tangent as the slope of the line in the unit circle. Since the line originates at #(0,0)#, its slope is
#tan(pi/3)=(sqrt3/2-0)/(1/2-0)=sqrt3#
This idea of #"slope"=(Deltay)/(Deltax)# is analogous to tangent because the sine values correlate to the #y# values of the ordered pair, and cosine with #x#, so remembering that #tan(x)=sin(x)/cos(x)# and that tangent is slope should be fairly intuitive.
