 My Math Forum Have no idea how to go about solving this triangle
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

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:
Attached Images Picture 2.png (53.3 KB, 118 views) September 8th, 2009, 08:08 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 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). 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: 20,921 Thanks: 2203 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�). 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: 20,921 Thanks: 2203 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. 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: 20,921 Thanks: 2203 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�. Tags idea, solving, triangle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mobel Number Theory 42 February 28th, 2014 10:56 PM covert Algebra 3 August 22nd, 2013 09:02 AM r-soy Physics 22 June 18th, 2012 11:16 AM manoj.ponnusamy Algebra 3 May 25th, 2011 05:18 AM therabidwombat Algebra 7 March 23rd, 2010 05:48 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      