Acoustic Guitar Construction

Gads. Half the battle is asking the right question, I think I have the attachment drawn correctly now.

I think this is how I will solve it
I would make the hidden triangle and use the tangent function on that new triangle as we want to know the opposite side and we know the adjacent and the angle.

So Tangent(90-a) = R/100 and Q = 50 - R

Doing some straight forward algebra and solving for Q you can type into Bing.com, google did not work,
50 - (Tangent (90-A) * 100) and that is Q; the angle needs to be between 63.5 and 90 degrees or you will go above or below your base.

For example A = 80 degrees

Got it! Thanks John. I used to love the hidden triangle method back in my geometry days. :-)

I have been lucky that I tutored both of my kids right up through calculus. I got to relearn everything at a high school pace, twice.

It can also be solved differently, using the cotangent function:

Q = cotangent A * [(tangent A * 50) - 100]

Basically, I simply extended Dave's triangle until it completed, then solved for the adjacent side of the triangle that completed his figure. Instead of the "hidden triangle" this approach could be called "the rest of the triangle".

John Link wrote:It can also be solved differently, using the cotangent function:

Q = cotangent A * [(tangent A * 50) - 100]

Basically, I simply extended Dave's triangle until it completed, then solved for the adjacent side of the triangle that completed his figure. Instead of the "hidden triangle" this approach could be called "the rest of the triangle".

Dave.jpg

If you draw it out is sort of uses a related hidden triangle that is on the other side of Q. But I see you got the slope and multiplied it by the run past Q thus getting Q.

I was shocked that you just can't type equations into google. For once bing wins

Yep, you are right. Neither triangle was visible in Dave's drawing.

What happened was I was offline doing my trig and when I came in to upload it discovered you had already put up your solution, which was a different path to the same result. Having done the work, I thought "what the heck" and recalculated mine using the 80 degrees.