Solve for the missing side lengths.V1045°A. Ou = 10/2, vV =10./33B. Ou-20v2, v =1033c. Ou = 20v2, v =10D. Ou = 10v2, v = 10
![Solve for the missing side lengthsV1045A Ou 102 vV 1033B Ou20v2 v 1033c Ou 20v2 v 10D Ou 10v2 v 10 class=](https://us-static.z-dn.net/files/d64/0a46949c930194bcc850341261491d99.png)
We have a right triangle, where we know that one of the angles (besides the right angle) has a measure of 45°.
Then, the other angle measure can be calculated as:
[tex]\begin{gathered} \alpha+45+90=180 \\ \alpha=180-90-45 \\ \alpha=45\degree \end{gathered}[/tex]Then, as the other angle measure is equal, we have an isosceles triangle.
Then, length v has to be equal to the side with length 10.
With the value of v we can calculate u with the Pythagorean theorem:
[tex]\begin{gathered} u^2=v^2+10^2 \\ u^2=10^2+10^2 \\ u^2=2\cdot10^2 \\ u=\sqrt[]{2}\cdot10 \\ u=10\sqrt[]{2} \end{gathered}[/tex]Answer: u = 10√2, v = 10