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September 8th, 2009, 07:34 AM   #1
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Have no idea how to go about solving this triangle

I have a right triangle problem that I just cannot figure out. Any point in the right direction would be greatly appreciated.

The problem in question:
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September 8th, 2009, 08:08 AM   #2
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Let's use A for the uppermost vertex of the triangle, B for the lower left vertes, and C for the lower right vertex, so that AC = 8 and BC = a. Let D be the point on AB such that the straight line through C and D is line h in your diagram.

Is your diagram intended to show that AB = 15 or that BD = 15?

If AB = 15, a = ?(15 - 8). Since the area of triangle ABC = 4a = 15h/2, it's then easy to find h.

The angles at A and B are respectively arctan(a/8) and arctan(8/a).
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September 8th, 2009, 08:41 AM   #3
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Re: Have no idea how to go about solving this triangle

Thanks a lot!

In response to your comment, I think the way CD intersects AB and then continues for a bit, im inclined to say 15 = BD and not AB. However, if this case proves unsolveable, I'll assume your method is correct.
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September 8th, 2009, 09:03 AM   #4
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If BD = 15, 15/a = a/?(a + 8) (by similar triangles or trigonometry). You can solve that for a by using the quadratic formula and hence find a. You can find h using h = ?(a - 15).
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September 9th, 2009, 04:55 AM   #5
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Re: Have no idea how to go about solving this triangle

Thanks again! The only question I have is what formulas are those? Could you tell me what part of trig. they could be found in, because I wanna be able to learn this for future reference.
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September 9th, 2009, 06:00 AM   #6
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If two triangles have the same angles, they are similar and have sides in the same proportions. This applies to the three triangles in the diagram. As they are right-angled, the definition of cosine can alternatively be used.

I also used the Pythagorean theorem and the quadratic formula.
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September 9th, 2009, 08:54 AM   #7
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Re: Have no idea how to go about solving this triangle

Ah, I see what you mean now. The two triangles created by BD are just scaled versions of one another, all the angles are the same. The one last bit I don't get, though, is the quadratic formula. How would I apply it to solve for a? I don't get what I would plug in for the other variables in the formula.
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September 9th, 2009, 09:49 AM   #8
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Squaring 15/a = a/?(a + 8) gives 225/a = a/(a + 64), which implies (a) - 225a - 14400 = 0, which is a quadratic equation in a.
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