Question

I need help with this question... it's about special triangles and I need to find y and z.. it should also not be a decimal.

Answer 1

To find z, consider the right-angled triangle at the botton in the diagram showm

`\begin{gathered} \sin 45\text{ = }(z)/(20) \\ z\text{ = 20 }\sin 45 \\ z\text{ = }20\text{ }*\frac{1}{\sqrt[]{2}} \\ z\text{ = }\frac{20}{\sqrt[]{2}} \\ z\text{ = }\frac{20}{\sqrt[]{2}}*\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ z\text{ = }\frac{20\sqrt[]{2}}{2} \\ z\text{ = 10}\sqrt[]{2} \end{gathered}`

Let the common base of both triangles be m

`\begin{gathered} \cos 45\text{ = }(m)/(20) \\ m\text{ = 20 }\cos 45 \\ m\text{ = }\frac{20}{\sqrt[]{2}} \\ m\text{ = }\frac{20}{\sqrt[]{2}}*\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ m\text{ = 10}\sqrt[]{2} \end{gathered}`

To find y:

`\begin{gathered} \tan 30\text{ = }(y)/(m) \\ \tan 30\text{ = }\frac{y}{10\sqrt[]{2}} \\ \frac{1}{\sqrt[]{3}}=\text{ }\frac{y}{10\sqrt[]{2}} \\ y\text{ = }\frac{10\sqrt[]{2}}{\sqrt[]{3}} \\ y\text{ = }\frac{10\sqrt[]{6}}{3} \end{gathered}`

To find x:

`\begin{gathered} \sin 30=(y)/(x) \\ (1)/(2)=\frac{10\sqrt[]{6}}{3}/ x \\ (1)/(2)=\frac{10\sqrt[]{6}}{3}*(1)/(x) \\ x\text{ = }\frac{20\sqrt[]{6}}{3} \end{gathered}`