# Given tanA=2, tanB=3, what is C if triangle ABC?

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You need to remember that the sum of internal angles in triangle is of `pi ` (`180^o` ) such that:

`A+B+C = pi =gt A+B = pi - C`

You should evaluate the tangent function of both sides such that:

`tan(A+B) = tan(pi - C)`

Using the formulas `tan(a+b) = (tan a + tan b)/(1 - tan a*tan b) ` and `tan(a-b) =(tan a- tan b)/(1+ tan a*tan b)` yields:

`(tanA + tan B)/(1 - tan A*tan B) = (tanpi - tan C)/(1 + tan a*tan C) `

Substituting 0 for `tan pi` yields:

`(tan A + tan B)/(1 - tan A*tan B) = -tan C`

The problem provides the values of `tan A` and `tan B` such that:

`(2 + 3)/(1 - 2*3) = -tan C`

`5/(-5) = -tan C =gt -tan C = -1 =gt tan C = 1 =gt C = 45^o`

**Hence, evaluating the measure of the angle C yields that `C = 45^o .` **

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