How do you find the exact value for #tan 120#?

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30
Jan 21, 2018

Answer:

#-sqrt3#

Explanation:

First of all #tan(angle) = (sin(angle)/cos(angle))#
So #tan(120) = sin(120)/cos(120)#
#sin(120)=sqrt(3)/2 and cos(120)=-1/2#
therefore #tan(120) = (sqrt(3)/2)/(-1/2)#
canceling out the #1/2# from the numerator and the denominator we get
#-sqrt(3)#
but remember #sqrt3# doesn't has any rational value .

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21
Jan 1, 2016

Answer:

#tan 120^@=-sqrt(3)#

Explanation:

#tan 120^@= sin120^@/cos120^@#

Let's evaluate #sin 120^@# first

#sin120^@= sin 2(60^@)#
#sin 120^@= 2sin 60^@cos60^@#
#sin 120^@= 2(sqrt3/2)(1/2)#
#sin 120^@= sqrt 3/2#

Since we found the value of #sin 120^@#, we can say that

#cos 120^@= -1/2#

Because #120^@# is in Quadrant 2.

Since #tan theta= sintheta/costheta, sqrt(3)/2*2#

(Dividing by 1/2 can be evaluated by doubling (multiplying by 2))

#:. tan 120^@=-sqrt(3)#

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11
Sep 18, 2017

Answer:

#tan(120) = - sqrt(3)#

Explanation:

#tan(120) = tan(2 * 60)#

We know that

#tan(2A) = (2tan(A))/(1 - tan^2(A))#

hence

#tan(2 * 60) = (2tan(60))/(1 - tan^2(60))#

Therefore,

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

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

# = (2 sqrt(3)) /( - 2)#

# = -sqrt(3)#

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