Question

How do you solve the right triangle ABC given Side a = 8, Side b = 10, Side c = 5?

Answer 1

`sqrt(8^2+5^2) != sqrt(10^2)`

Therefore
`color(white)("XXXX")`a triangle with sides `a=8`, `b=10`, and `c=5` is not a right triangle.

With the given sides we can determine the angles using the Law of Cosines. For example
`color(white)("XXXX")` `cos(C) = (a^2+b^2-c^2)/(2ab)`

`color(white)("XXXX")` `color(white)("XXXX")` `= (64+100-25)/(160) = 139/160`

`color(white)("XXXX")` `C = arccos(139/160)`

`color(white)("XXXX")` `color(white)("XXXX")` `= 0.518124 " radians " = 29.6863^@`