Which angles?

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
need2know
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Which angles?

A slightly odd problem that I can't seem to think my way through.

I have a non-oblique triangle.

I know the base and the height, and I know the difference between the lengths of the other two sides. How would I calculate the lengths of those two sides?

Thanks!

buddy
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Re: Which angles?

I have a non-oblique triangle.

I know the base and the height, and I know the difference between the lengths of the other two sides. How would I calculate the lengths of those two sides?
non-oblique means right so use the Pythagorean Theorem
I don't know why you need angles?

need2know
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Re: Which angles?

Sorry, ignore the bit about oblique.

Basically I really want a formula that I can use to plot the location of the peak (i.e. corner at the top of the height line), given a fixed base, a constant difference between the lengths of the other two sides, and with the height known but varying.

Thanks

nona.m.nona
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Re: Which angles?

It might help if you posted the exact text of the homework exercise, so that the parameters might better be understood.

need2know
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Re: Which angles?

Two Cartesian points a,b and c,d.

Given this equation:

e = abs(sqrt((a - x)2 + (b - y)2) - sqrt((c - x)2 + (d - y)2))

How can I solve for y if all other variables are known.

Hope that makes it clearer. I assumed a trigonometric solution was the way to go.

nona.m.nona
Posts: 288
Joined: Sun Dec 14, 2008 11:07 pm
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Re: Which angles?

Given this equation:

e = abs(sqrt((a - x)2 + (b - y)2) - sqrt((c - x)2 + (d - y)2))

How can I solve for y if all other variables are known.

Hope that makes it clearer. I assumed a trigonometric solution was the way to go.
I'm not sure how trigonometric functions would apply to the solution methods for this algebraic equation. The solution would likely start by squaring both sides (which will eliminate the absolute value), simplifying, isolating the remaining radical, and then squaring again.