# Basic trig question regarding definition of sin

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?

• When I put $\sin(63.43^\circ)$ into a calculator, I obtain $\approx 0.89$. You accidentally used radian mode, which gave you $\sin(63.43) \approx 0.56$. – N. F. Taussig Aug 25 '19 at 2:39
• @N.F.Taussig omg! Totally forgot about radian and degrees! Been a long time since I did math. Thanks! – Robben Aug 25 '19 at 2:40
• Check the mode. Switch from radians to degrees. – N. F. Taussig Aug 25 '19 at 2:41
• @Robben A calculator is much more rarely wrong than its operator. Check the mode you put it in. Radians or degrees? – Deepak Aug 25 '19 at 2:42
• @N. F. Taussig It's not radians vs degrees, it's the need to use the inverse sine function. – 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=$$\sqrt{0.5^2+1^2}=1.118033989$$.
$$\tan\theta=\frac{opposite}{adjacent}=\frac{1}{0.5}=2\implies \arctan 2=1.107148718^{radians}=63.43^\circ$$.
$$\sin\theta=\frac{1}{1.118033989}=0.8944\implies\theta=\arcsin0.8944=1.107148718^{radians}=63.43^\circ$$
$$\cos\theta=\frac{0.5}{1.118033989}=0.4472\implies\theta=\arccos0.4472=1.107148718^{ radians}=63.43^\circ$$