• Feb 9th 2009, 11:43 AM
jamm89
Trionometry with a bit of pythag
Hi i am new here.
Can anyone point me in the right direction on this one please
http://i75.photobucket.com/albums/i3...scan0002-1.jpg
b=15.7

I have gone as far as splitting the 2 triangles up and labeling the 2
24.3² + DB = BC²
15.7² + DB² = AB² (DB being the vertical unknown side)

I then did AB + BC = 42 so thefore AB=42-BC and BC=42-AB

I substituted these values in to the equation i came up with
24.3² + DB = (42-AB)
15.7² + DB² = (42-BC)

I then tryed subtracting these away from each other as if a symlytaineous equation. (dont know if that was the right way to go about it)

Thanks
• Feb 10th 2009, 01:39 AM
earboth
Quote:

Originally Posted by jamm89
Hi i am new here.
Can anyone point me in the right direction on this one please
...

b=15.7

...

I've modified your sketch a little bit. (see attachment)

You've got two right triangles. Use Pythagorean theorem:

$\begin{array}{rcl}b^2+d^2& = &a^2 \\(40-b)^2+d^2&=&(42-a)^2\end{array}$

Expand the brackets and subtract the first equation from the second one:

$1600-80b = 1764-84a$

Now plug in the given value for b and solve for a. I've got a = 16.9. Thus d = 6.25.

You now have all values to calculate the angles at B.

I asume that you know how to add forces and that you need a parallelogram to get all partial forces.