# A Triangle

• May 23rd 2008, 05:24 PM
ty2391
A Triangle
This question is related to a vector/force question, but I just need help with finding two angles:

Given the following triangle:

http://img511.imageshack.us/img511/8786/88666773wl2.jpg

and knowing that a+b = 14. How can you figure out the length of a and b?
• May 23rd 2008, 07:58 PM
Isomorphism
Quote:

Originally Posted by ty2391
This question is related to a vector/force question, but I just need help with finding two angles:

Given the following triangle:

http://img511.imageshack.us/img511/8786/88666773wl2.jpg

and knowing that a+b = 14. How can you figure out the length of a and b?

Well I can clearly see a right angled triangle. So i think the answer is a = 8 and b = 6 :)
• May 23rd 2008, 08:01 PM
ty2391
Sorry, the diagram may be misleading, im trying to say that the 6 and 4 split the top into two sections. (thanks for your reply though)

Anyway, I figured it out I believe, can someone confirm that: a=54/7 and b=44/7?
• May 23rd 2008, 08:07 PM
topsquark
Quote:

Originally Posted by ty2391
This question is related to a vector/force question, but I just need help with finding two angles:

Given the following triangle:

http://img511.imageshack.us/img511/8786/88666773wl2.jpg

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
$a^2 = 6^2 + c^2 = 36 + c^2$
and
$b^2 = 4^2 + c^2 = 16 + c^2$

Then
$a^2 - b^2 = 20$

But
$a^2 - b^2 = (a + b)(a - b)$
and we know that $a + b = 14$. Thus
$14(a - b) = 20$

$a - b = \frac{10}{7}$

Thus we have the system of equations
$a + b = 14$

$a - b = \frac{10}{7}$

This has a solution of
$a = \frac{54}{7}$
and
$b = \frac{44}{7}$

(Which are at least close to Isomorphism's a = 8 and b = 6.)

-Dan