# Thread: A triangulation help area needed! ASAP

1. ## A triangulation help area needed! ASAP

hi thanks for looking at this.

i have a triangle with angles ABC

A=57 deegrees
B=45 deegrees
C= not given

and one length between AB of 14metres

I need to calculate the area within the triangle?

can someone work it out and show me how?

thanks

james

2. Originally Posted by jaynlisa08
hi thanks for looking at this.

i have a triangle with angles ABC

A=57 deegrees
B=45 deegrees
C= not given

and one length between AB of 14metres

I need to calculate the area within the triangle?

can someone work it out and show me how?

thanks

james
No offense, but did you even try attempting this problem?
It's obvious $\displaystyle C = 78$ degrees.

Construct line AM perpendicular to BC.

$\displaystyle Sin(45) = \frac{AM}{AB} = \frac{AM}{14}$

$\displaystyle Sin(45) = \frac{ \sqrt{2} }{2}$

$\displaystyle AM = 7 \sqrt{2}$

Now use the Sin Rule (Think that's what it's called )

$\displaystyle \frac{Sin C}{c} = \frac{Sin A}{a}$

$\displaystyle \frac{Sin \ 78}{14} = \frac{Sin \ 57}{a}$

$\displaystyle a = BC = 12,0037$

Now the area is given by:

$\displaystyle \frac{1}{2} \times BC \times AM$

3. actually its the sine law