I have a very basic trig question.

I have a right angle triangle. The triangle $y$-axis has size $0.1$ meters and the $x$-axis has $0.05$ meters. Now using the definition of tangent, I have $\tan \theta = \frac{opposite}{adjecent}$ therefore the angle for the triangle is $63.43$. Now, $\sin \theta =\sin(63.43) =0.56 $. Using the definition of $\sin$ I have that $\sin\theta=\frac{opposite}{hypotenuse}= \frac{.1}{\sqrt{(.1)^2+(.05)^2}} = 0.89$ which doesn't equal $\sin(63.43)$.

Not sure what I am doing wrong?

  • 4
    $\begingroup$ When I put $\sin(63.43^\circ)$ into a calculator, I obtain $\approx 0.89$. You accidentally used radian mode, which gave you $\sin(63.43) \approx 0.56$. $\endgroup$ – N. F. Taussig Aug 25 '19 at 2:39
  • $\begingroup$ @N.F.Taussig omg! Totally forgot about radian and degrees! Been a long time since I did math. Thanks! $\endgroup$ – Robben Aug 25 '19 at 2:40
  • $\begingroup$ Check the mode. Switch from radians to degrees. $\endgroup$ – N. F. Taussig Aug 25 '19 at 2:41
  • 2
    $\begingroup$ @Robben A calculator is much more rarely wrong than its operator. Check the mode you put it in. Radians or degrees? $\endgroup$ – Deepak Aug 25 '19 at 2:42
  • $\begingroup$ @N. F. Taussig It's not radians vs degrees, it's the need to use the inverse sine function. $\endgroup$ – poetasis Aug 25 '19 at 4:24

You forgot to use $\arcsin0.89$. Let's be sure of terms:

adjacent=$x=0.5\quad$opposite=$y=1\quad$ hypotenuse=$\sqrt{0.5^2+1^2}=1.118033989$.

$\tan\theta=\frac{opposite}{adjacent}=\frac{1}{0.5}=2\implies \arctan 2=1.107148718^{radians}=63.43^\circ$.


$\cos\theta=\frac{0.5}{1.118033989}=0.4472\implies\theta=\arccos0.4472=1.107148718^{ radians}=63.43^\circ $

| cite | improve this answer | |

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.