<s>[INST] If $\tan \theta = 3\frac{15}{16}$, then find $\sin \theta$ If $\tan \theta = 3\cfrac{15}{16}$, then find $\sin \theta$.
 [/INST] Let us rewrite $3 \frac{15}{16}$ as an improper fraction. So we get $3 \dfrac{15}{16} = \dfrac{16\cdot3+15}{16} = \dfrac{63}{16}$
We know that $\tan(\theta) = \dfrac{o}{a}$, where $o$ and $a$ are the opposite and adjacent sides of $\theta$. We also know through the Pythagorean theorem that the hypotenuse is $\sqrt{o^2+a^2} = \sqrt{63^2+16^2} = 65$. Recall that:
$$\sin(\theta) = \frac{o}{h}$$ where $h$ is the hypotenuse and $o$ is the opposite side. Subbing in the values we obtained and get:
$$\sin(\theta) = \frac{63}{65}$$
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