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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?

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

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

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

$\sin\theta=\frac{1}{1.118033989}=0.8944\implies\theta=\arcsin0.8944=1.107148718^{radians}=63.43^\circ$

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

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