# Solve: $\tan(2\theta-36^\circ) = \sqrt{8}$

$\tan(2\theta-36^\circ) = \sqrt{8}$ in degrees.

I tried making $\sqrt{3}$ into $60$ degrees, and then the answer was $47$ degrees but I don't think that is right.

• It seems that some very clever answers were based on the original problem statement. Try updating the second sentence (referencing $\sqrt{3}$, $60$ degrees, and $47$ degrees), to reflect the new problem statement as it seems that this is becoming a pentagon problem, and not one based on an equilateral triangle. – John Joy Apr 24 '15 at 13:55

If you let $2\theta-34 = \beta$

Then we know: $tan(\beta)=\sqrt3= \frac{opposite}{adjacent}$

And from the attached diagram we see the angle should be: $\beta=60 \implies \theta=47$

• This is incorrect. $\theta=32$ is not a solution of OP's equation. – MPW Apr 23 '15 at 0:44
• Yes, now quite correct. Nice diagram -- how did you generate it? – MPW Apr 23 '15 at 0:47
• Credit goes to Google's image search. – CivilSigma Apr 23 '15 at 0:50
• Haha! Clever! The underlying triangle is indeed probably extremely commonly illustrated. That's a neat way of thinking about it. +1. – MPW Apr 23 '15 at 0:51

Hint: First solve $\tan x =\sqrt{3}$ for $x$, then solve $x=2\theta -34$ for $\theta$ (using the value for $x$ you found from the first equation).

And your answer is correct. There are also other solutions, since the first equation I wrote has multiple solutions (the function is periodic).

• The question says theta is real numbers of degrees, does that change anything? – Elle Apr 23 '15 at 3:47
• No, just that when you solve the first equation, you will get values of $x$ that are degrees, not radians. It is implicit that $x$ is a number of degrees since that's how you stated the problem. – MPW Apr 23 '15 at 4:06

Tangent has period $\pi$ and is one-to-one within each period. Thus, we know that $$2\theta-34^\circ\equiv60^\circ\pmod{180^\circ}$$ Therefore, $$\theta\equiv47^\circ\pmod{90^\circ}$$