Answer
The triangle in the figure 4 is correct .
Reason
by using the trignometric identity
[tex]tan\theta = \frac{Perpendicular}{Base}[/tex]
Thus
[tex]tan 30^{\circ} = \frac{Perpendicular}{Base}[/tex]
[tex]tan 30 ^{\circ} =\frac{1}{\sqrt{3}}[/tex]
Now in the figure (1)
[tex]tan 30^{\circ} = \frac{13.9}{8}[/tex]
[tex]\frac{1}{\sqrt{3}} = \frac{13.9}{8}[/tex]
on simplify
[tex]0.5774 \neq 1.7375[/tex]
thus side length measures in the figure (1) is not correct .
Now in the figure (2)
[tex]tan30^{\circ} =\frac{16}{8} \\ \frac{1}{\sqrt{3}} =\frac{16}{8} \\ 0.577 \neq 2[/tex]
thus side length measures in the figure (2) is not correct .
Now in the figure (3)
[tex]tan 30 ^{\circ} = \frac{8}{16} \\ \frac{1}{\sqrt{3}} =\frac{8}{16} \\ 0.577\neq0.5[/tex]
thus side length measures in the figure (3) is not correct .
Now in the figure (4)
[tex]tan30^{\circ} = \frac{8}{13.9} \\ \frac{1}{\sqrt{3}} =\frac{8}{13.9}\\ 0.577 = 0.576(approx)[/tex]
Therefore the figure (4) is correct triangle has side length measures that could be correct
Hence proved
