What is the unit circle value of tan 120, 135, and 150 degrees?

Use #tan theta = sin(theta)/cos(theta)#

From a trig circle or a #30^@-60^@-90^@# triangle in the second quadrant:
#tan 120^@ = (sqrt(3)/2)/(-1/2) = sqrt(3)/2 *-2/1 = -sqrt(3)#

From a trig circle or a #45^@-45^@-90^@# triangle in the second quadrant:
#tan 135^@ = (sqrt(2)/2)/(-sqrt(2)/2) = sqrt(2)/2 * -2/sqrt(2) = -1#

From a trig circle or a #30^@-60^@-90^@# triangle in the second quadrant:
#tan 150^@ = (1/2)/(-sqrt(3)/2) = 1/2 * -2/sqrt(3) = -1/sqrt(3) = -1/sqrt(3) * sqrt(3)/sqrt(3) = -sqrt(3)/3#

Let's use the unit circle to find the values

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#color(blue)(tan(120^circ)#

We have the values of #sin(120^circ) and cos(120^circ)#

So, use the identity

#color(brown)(tan(theta)=( sin(theta))/(cos(theta))#

#rarrtan(120^circ)=(sin(120^circ))/(cos(120^circ))#

#rarrtan(120^circ)=(sqrt(3)/2)/(-1/2)#

#rarrtan(120^circ)=sqrt(3)/2*-2/1#

#rarrtan(120^circ)=-cancel2sqrt(3)/cancel2#

#color(blue)(rArrtan(120^circ)=-sqrt(3)#

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#color(orange)(tan(135^circ)#

#color(brown)(tan(theta)=( sin(theta))/(cos(theta))#

#rarrtan(135^circ)=(sin(135^circ))/(cos(135^circ))#

#rarrtan(135^circ)=(sqrt(2)/2)/(-sqrt(2)/2)#

#rarrtan(135^circ)=2/sqrt2*-2/sqrt(2)#

#rarrtan(135^circ)=-cancel((2sqrt2)/(2sqrt2)#

#color(orange)(rArrtan(135^circ)=-1#

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#color(purple)(tan(150^circ)#

#color(brown)(tan(theta)=( sin(theta))/(cos(theta))#

#rarrtan(150^circ)=(sin(150^circ))/(cos(150^circ))#

#rarrtan(150^circ)=(1/2)/(-sqrt(3)/2)#

#rarrtan(150^circ)=1/2*-2/sqrt(3)#

#rarrtan(150^circ)=-cancel2/(cancel2sqrt(3))#

#rarrtan(150^circ)=-1/3#

Here is the most important part. The denominator is an irrational number, so multiplu both numerator and denominator by #sqrt3#

#rarrtan(150^circ)=-1/3*(sqrt(3)/sqrt(3))#

#color(purple)(rArrtan(150^circ)=-(sqrt(3))/3#

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Hope this helps!!! :)