a) 5
b) 10
c) 5
d) 5√3
Explanation:From the left triangle:
hypotenuse = 5√2
opposite = a
adjacent = c
angle = 45°
To get a, we will use sine ratio:
[tex]\begin{gathered} \sin \text{ 45}\degree\text{ = }\frac{opposite}{hypotenuse} \\ \sin \text{ 45}\degree\text{ = }\frac{a}{5\sqrt[]{2}} \\ 5\sqrt[]{2}\text{ (sin45}\degree)=a_{} \\ 5\sqrt[]{2}\text{ }\times\text{(}\frac{\sqrt[]{2}}{2})\text{ = a} \\ \frac{5(2)}{2}\text{ = a} \\ a\text{ = 5} \end{gathered}[/tex]To get c, we will use cosine ratio:
[tex]\begin{gathered} \cos \text{ 45}\degree\text{ = }\frac{adjacent}{\text{hypotenuse}} \\ \cos 45\degree\text{ = }\frac{c}{5\sqrt[]{2}} \\ 5\sqrt[]{2}\text{ (cos 45}\degree)\text{ = c} \\ 5\sqrt[]{2}\text{ (}\frac{\sqrt[]{2}}{2})\text{ = c} \\ \frac{5(2)}{2}\text{ = c} \\ c\text{ = 5} \end{gathered}[/tex]From right triangle:
angle = 30°
To get b, we will apply sine ratio:
[tex]\begin{gathered} \sin \text{ 30}\degree\text{ = }\frac{opposite}{\text{hypotenuse}} \\ \sin \text{ 30}\degree\text{ = }\frac{a}{b} \\ \sin \text{ 30}\degree\text{ = }\frac{5}{b} \\ b(\sin \text{ 30}\degree\text{) = 5} \\ b(\frac{1}{2})\text{ = 5} \\ b\text{ = 2(5) } \\ b\text{ = 10} \end{gathered}[/tex]To get d, we will apply cosine ratio:
[tex]\begin{gathered} \cos \text{ 30}\degree\text{ = }\frac{adjacent}{\text{hypotenuse}} \\ \cos \text{ 30}\degree\text{ = }\frac{d}{b} \\ \cos \text{ 30}\degree\text{ = }\frac{d}{10} \\ 10(\cos \text{ 30}\degree)\text{ = d} \\ 10\times\frac{\sqrt[]{3}}{2}\text{ = d} \\ d\text{ = 5}\sqrt[]{3} \end{gathered}[/tex]