# Proof in a triangle.

• Feb 6th 2010, 03:37 AM
BabyMilo
Proof in a triangle.
Proof AB=3.6sin0.2

i've done.

$\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$

$\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$

but duno how to derive the answer.

thanks!
• Feb 6th 2010, 03:43 AM
Prove It
Quote:

Originally Posted by BabyMilo
Proof AB=3.6sin0.2

i've done.

$\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$

$\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$

but duno how to derive the answer.

thanks!

• Feb 6th 2010, 04:00 AM
BabyMilo
Quote:

Originally Posted by Prove It

Still I dont understand how i would dervie the required.

$\displaystyle a=\sqrt{2*1.8^2-2*1.8*1.8cos0.4}$

They are not whole numbers so how would i find the required. and since this is in cos.

thanks!
• Feb 6th 2010, 05:35 AM
SiriusAlpha
Is it just,

$\displaystyle a=\sqrt{2*1.8^2-2*1.8*1.8cos0.4}$
$\displaystyle a=\sqrt{2*3.24-3.6*1.8(0.921)}$
$\displaystyle a=\sqrt{2*3.24-3.6*1.65}$

Or is there something I'm not getting D:
• Feb 6th 2010, 06:29 AM
BabyMilo
Quote:

Originally Posted by BabyMilo
Proof AB=3.6sin0.2

i've done.

$\displaystyle \frac{a}{sin0.4} = \frac{1.8}{sin1.37}$

$\displaystyle \rightarrow a=\frac{1.8sin0.4}{sin1.37}$

but duno how to derive the answer.

thanks!

Solve it nevermind.

cut the triangle into half.

so it's 1.8sin0.2 x 2
=3.6sin0.2 as required.