[tex]\bf cos(\theta)=\cfrac{3}{5}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{5^2-3^2}=b
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\pm\sqrt{16}=b\implies \pm4=b[/tex]
so, plus or minus? well, the angle is in the 1st quadrant, and on that quadrant, both cosine and sine are positive, so 4 = b
[tex]\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{4}{3}[/tex]
