Respuesta :
well, we know the sine, and we also know that we're on the II Quadrant, let's recall that on the II Quadrant sine is positive whilst cosine is negative.
[tex]\bf sin^2(\theta)+cos^2(\theta)=1~\hspace{10em} tan(\theta )=\cfrac{sin(\theta )}{cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ sin^2(a)+cos^2(a)=1\implies cos^2(a) = 1-sin^2(a) \\\\\\ cos^2(a) = 1-[sin(a)]^2\implies cos^2(a) = 1-\left( \cfrac{3}{4} \right)^2\implies cos^2(a) = 1-\cfrac{3^2}{4^2} \\\\\\ cos^2(a) = 1-\cfrac{9}{16}\implies cos^2(a) = \cfrac{7}{16}\implies cos(a)=\pm\sqrt{\cfrac{7}{16}}[/tex]
[tex]\bf cos(a)=\pm\cfrac{\sqrt{7}}{\sqrt{16}}\implies cos(a)=\pm\cfrac{\sqrt{7}}{4}\implies \stackrel{\textit{on the II Quadrant}}{cos(a)=-\cfrac{\sqrt{7}}{4}}\\\\[-0.35em]~\dotfill\\\\tan(a)=\cfrac{sin(a)}{cos(a)}\implies tan(a)=\cfrac{~~\frac{3}{4}~~}{-\frac{\sqrt{7}}{4}}\implies tan(a)=\cfrac{3}{4}\cdot \cfrac{4}{-\sqrt{7}}\\\\\\tan(a)=-\cfrac{3}{\sqrt{7}}\implies \stackrel{\textit{rounded up}}{tan(a) = -1.13}[/tex]
