Trig Question

[au5233]

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Hey Guys,

I'm working on my favorite subject math (sarcasm) and came across a problem. I'm using the Kaplan book and the solution has:

tanθ = 3
θ = arctan 3

θ = 71.6 degrees


Ok so when I was doing this problem, I made it down to the tanθ= 3 but I have no idea how to do the rest.

Can someone break it down for me, how to find the arctan of something?

 

[Streetwolf]

Okay well... it's just an educated guess you have to make. There's no identity for this one.

I'm working in degrees here...

tan(0) = 0
tan(30) = sqrt(3) / 3 = 0.577 (sin(30)/cos(30))
tan(45) = 1
tan(60) = sqrt(3) = 1.732
tan(90) = (goes to infinity)

So it's somewhere between 60 and 90 degrees. Unless you know any values of sin and cos between 60 and 90 degrees, that's about the best you can do.

Kaplan is notorious for making you use a calculator. There won't be anything quite like this on the real test.

 

[jay47]

Yeah, he knows what he is talking about, there won't be any trig this difficult on the test. Streetwolf is just about a genius in this section so I would take his advice seriously. If they put a trig question on the test that requires an identity, I would think that it would require very few steps to rearrange to find the answer.

 

[au5233]

HUGE help guys, thank you.

I thought that's what u would have to do...kind of estimate based on special triangles and the tan graph. Stupid kaplan...they just jump from one step to the next...grr.

lol best of luck studying.