Answer:
[tex]d=\frac{4}{\sqrt[]{3}},c=8\sqrt[]{2}[/tex]
Explanation:
Use the trigonometric ratio,
[tex]\sin \theta^{}=\frac{Opposite\text{ side}}{Hypotenuse}[/tex]
From the first right angled triangle,
[tex]\begin{gathered} \sin 60^o=\frac{2}{d} \\ d=\frac{2}{\sin60^o} \\ =\frac{2}{\frac{\sqrt[]{3}}{2}} \\ =\frac{4}{\sqrt[]{3}} \end{gathered}[/tex]
Similarly, from the second right angled triangle,
[tex]\begin{gathered} \sin 45^o=\frac{8}{c} \\ c=\frac{8}{\sin 45^o} \\ =\frac{8}{\frac{1}{\sqrt[]{2}}} \\ =8\sqrt[]{2} \end{gathered}[/tex]
