This question is related to a vector/force question, but I just need help with finding two angles:
Given the following triangle:
and knowing that a+b = 14. How can you figure out the length of a and b?
Call the clearly vertical leg (that obviously is intended to meet the side of length 10 at a right angle) c.
Then we know that
$\displaystyle a^2 = 6^2 + c^2 = 36 + c^2$
and
$\displaystyle b^2 = 4^2 + c^2 = 16 + c^2$
Then
$\displaystyle a^2 - b^2 = 20$
But
$\displaystyle a^2 - b^2 = (a + b)(a - b)$
and we know that $\displaystyle a + b = 14$. Thus
$\displaystyle 14(a - b) = 20$
$\displaystyle a - b = \frac{10}{7}$
Thus we have the system of equations
$\displaystyle a + b = 14$
$\displaystyle a - b = \frac{10}{7}$
This has a solution of
$\displaystyle a = \frac{54}{7}$
and
$\displaystyle b = \frac{44}{7}$
(Which are at least close to Isomorphism's a = 8 and b = 6.)
-Dan
EDIT: Just saw your post ty2391. Obviously I agree with your answer. Good job.