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September 8th, 2009, 07:34 AM  #1 
Newbie Joined: Sep 2009 Posts: 4 Thanks: 0  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: 
September 8th, 2009, 08:08 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,485 Thanks: 2041 
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 = ?(15²  8²). 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). 
September 8th, 2009, 08:41 AM  #3 
Newbie Joined: Sep 2009 Posts: 4 Thanks: 0  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. 
September 8th, 2009, 09:03 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,485 Thanks: 2041 
If BD = 15, 15/a = a/?(a² + 8²) (by similar triangles or trigonometry). You can solve that for a² by using the quadratic formula and hence find a. You can find h using h = ?(a²  15²).

September 9th, 2009, 04:55 AM  #5 
Newbie Joined: Sep 2009 Posts: 4 Thanks: 0  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.

September 9th, 2009, 06:00 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,485 Thanks: 2041 
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 rightangled, the definition of cosine can alternatively be used. I also used the Pythagorean theorem and the quadratic formula. 
September 9th, 2009, 08:54 AM  #7 
Newbie Joined: Sep 2009 Posts: 4 Thanks: 0  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.

September 9th, 2009, 09:49 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,485 Thanks: 2041 
Squaring 15/a = a/?(a² + 8²) gives 225/a² = a²/(a² + 64), which implies (a²)²  225a²  14400 = 0, which is a quadratic equation in a².


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