2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – John Joy Apr 24 '15 at 13:55
2
$\begingroup$

The special angle triangle!

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: $\beta=60 \implies \theta=47$

$\endgroup$
4
  • 1
    $\begingroup$ This is incorrect. $\theta=32$ is not a solution of OP's equation. $\endgroup$ – MPW Apr 23 '15 at 0:44
  • $\begingroup$ Yes, now quite correct. Nice diagram -- how did you generate it? $\endgroup$ – MPW Apr 23 '15 at 0:47
  • $\begingroup$ Credit goes to Google's image search. $\endgroup$ – CivilSigma Apr 23 '15 at 0:50
  • $\begingroup$ Haha! Clever! The underlying triangle is indeed probably extremely commonly illustrated. That's a neat way of thinking about it. +1. $\endgroup$ – MPW Apr 23 '15 at 0:51
1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ The question says theta is real numbers of degrees, does that change anything? $\endgroup$ – Elle Apr 23 '15 at 3:47
  • $\begingroup$ 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. $\endgroup$ – MPW Apr 23 '15 at 4:06
0
$\begingroup$

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, $$ \theta\equiv47^\circ\pmod{90^\circ} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.