The triangles is the bigger triangle are both right angled
Considering the triangle by the left:
when angle = 60°
opposite = d
hypotenuse = 42
To get d, we will use sine ratio:
[tex]\begin{gathered} \sin \text{ 60}\degree\text{ = }\frac{opposite}{hypotenuse} \\ \sin \text{ 60}\degree\text{= }\frac{d}{42} \\ d\text{ = 42 }\times\text{ sin 60}\degree\text{ } \\ In\text{ radical form sin60}\degree\text{ = }\frac{\sqrt[]{3}}{2} \\ d\text{ =42 }\times\frac{\sqrt[]{3}}{2} \\ d\text{ = }21\text{ }\sqrt[]{3} \end{gathered}[/tex]To get a, we will use cosine ratio:
[tex]\begin{gathered} \cos \text{ 60}\degree\text{ = }\frac{adjacent}{hypotenuse} \\ \cos \text{ 60}\degree\text{in radical form = }\frac{1}{2} \\ \text{adjacent = a} \\ \cos \text{ 60}\degree\text{ = }\frac{a}{42} \\ a\text{ = 42(}\cos \text{ 60}\degree)\text{ }=\text{ 42(}\frac{1}{2}) \\ a\text{ = }21 \end{gathered}[/tex]Considering the triangle by the right:
opposite = d = 21√3
angle = 30°
hypotenuse = c
To get c, we will use sine ratio:
[tex]\begin{gathered} \sin \text{ 30}\degree\text{ = }\frac{opposite}{hypotenuse} \\ \sin \text{ 30}\degree\text{ = }\frac{21\surd3}{c} \\ c\text{ = }\frac{21\surd3}{\sin \text{ 30}\degree}\text{ ; }\sin \text{ 30}\degree\text{ = 1/2 } \\ c\text{ = }\frac{21\surd3}{\frac{1}{2}}=\text{ }21\surd3\text{ }\times\text{ 2} \\ c\text{ = 42 }\sqrt[]{3} \end{gathered}[/tex]b = adjacent
To get b, we will use cosine ratio:
[tex]\begin{gathered} \cos \text{ 30}\degree\text{ = }\frac{adjacent}{\text{hypotenuse}}\text{= }\frac{b}{c} \\ \cos \text{ 30}\degree\text{ }=\frac{b}{42\text{ }\sqrt[]{3}} \\ \cos \text{ 30}\degree\text{ in radical form = }\frac{\sqrt[]{3}}{2} \\ b\text{ = }42\text{ }\sqrt[]{3}\text{ }\times\text{ }\cos \text{ 30}\degree\text{ = }42\text{ }\sqrt[]{3}\text{ }\times\frac{\sqrt[]{3}}{2} \\ b\text{ = }\frac{42(3)}{2} \\ b\text{ = 63} \end{gathered}[/tex]