Similar triangles: find length of B in terms of X and theta

[MarkSA]

Hello,

Here is a problem I have:

I think it's dealing with similar triangles, but i'm not sure how to write B in terms of X and theta. If X was the entire length of big triangle, it would not be a problem, but it's only a side of the inner triangle. All i've really determined is that the angle opposite X is also theta, and I suppose I could say the other angle is 90degrees - theta.

Any suggestions?

 

[MarkSA]

Re: Similar triangles

I got an answer of B = x/(sin(theta)tan(theta)) by assigning the side in the middle of the big triangle the value of 'a' and then solving both inner triangles for a. I then set them equal and then solved for B.

sin(theta) = a/B
tan(theta) = X/a
a = x/tan(theta)
a = Bsin(theta)
x/tan(theta) = Bsin(theta)
B = x/(tan(theta)sin(theta))

But I think there must be an easier way to do this one.. is there?

 

[pka]

Re: Similar triangles

\(\displaystyle \tan (\theta ) = \frac{Y}{B}\,\& \,\sin (\theta ) = \frac{X}{Y}\)
\(\displaystyle X = Y\sin (\theta ) = B\tan (\theta )\sin (\theta )\)

 

[MarkSA]

Thanks, so the way I did it was correct?

I was a little surprised because this is a review problem for our physics class' "basic math minitest". All other similar review problems were much easier to do than this one.