solve: tan(`theta` -20`degrees` )=`sqrt(3)`

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To solve: `tan(theta-20) = sqrt3` it is important to recognize that `sqrt3` is a ratio in the special angles triangle 30 - 60 - 90 where we have

`2:1:sqrt3` where 2 is the longest side and is therefore the hypoteneuse.

`tan(theta-20) = (opp)/(adj) = sqrt3/1` (that is; the ratio of the other two sides) We add the denominator of 1 to show the ratio. In terms of the special angles:

`tan 60 = sqrt3/1`

`therefore tan (80 - 20) = tan 60 = sqrt3/1`

`therefore theta=80` It could also exist in Quad III as `(-sqrt3)/(-1)=sqrt3/1` which is`(180 + theta)` as we have a positive ratio which renders `180+80 = 260`

**Ans: therefore `theta` = 80 degrees or `theta` = 260 degrees**

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