Finding Sides and AnglesDate: 01/22/2001 at 01:22:36 From: john yee Subject: Geometry Dear Dr. Math, I have a real-life question that goes beyond my high school training or I've just simply forgotten how to do it. I suspect this is a geometry question. Please help if you can. Thank you. This is a right triangle of which one side and the right angle are known. Also given is an elevation value that is not the true length of any of the triangle's sides. From a horizontal and vertical intersection begins a 5/32 long leg of a right triangle 10 degrees upward from the horizontal line. At the distance of 5/32 another leg upward is perpendicular to the 5/32 line and ends on a line parallel and 2 1/2 away from the horizontal line. This point and the horizontal vertical intersection determine the hypotenuse. Can you determine the length of the hypotenuse and an acute interior angle of this right triangle? Can you show me how to do this and identify the subclass of geometry where I may learn more about it? Thanks again. Date: 01/22/2001 at 10:52:00 From: Doctor Rob Subject: Re: Geometry Thanks for writing to Ask Dr. Math, John. This is a problem in trigonometry. There is enough information to find all the sides and angles of the triangle. Here is a rough diagram: C ---o---------------------------------- F horizontal line |\ 80 degrees |\\ | \ | \ \ | y\ | \ \ | \ | \ \ | \ | \ \ | \ | \ \ | \ | \ \ | \ | \ \ | \ | \x \ |5/2 \ | \ \ | \ | \ \ | \ | \ \ | \ | \ \ | 5/32 _,o B | \ _,-' | _,-' 10 degrees ---o--------------o------------------ E horizontal line D A The triangle described is ABC, <B = 90 degrees, <BAE = 10 degrees. Length AB = 5/32 and length CD = 5/2. Then <BCD = 10 degrees (since the sides of <BCD are perpendicular to the sides of <BAE). Let length AC, which you seek, be x, and <ACB = y. Then, looking at right triangle ACD, cos(10-y) = (5/2)/x = 5/(2*x), and looking at right triangle ABC, sin(y) = (5/32)/x = 5/(32*x). This forms a system of two equations in two unknowns x and y, which you need to solve. The solution you seek will have x > 5/2 and 0 < y < 10 degrees. I would solve this system by using the trigonometric identities cos(a-b) = cos(a)*cos(b) + sin(a)*sin(b), sin^2(a) + cos^2(a) = 1. The first equation above and the first identity give cos(10)*cos(y) + sin(10)*sin(y) = 5/(2*x). Substituting for sin(y) its equal from the second equation above gives cos(10)*cos(y) + sin(10)*5/(32*x) = 5/(2*x). Solving for cos(y), you get cos(y) = [5/(2*x) - sin(10)*5/(32*x)]/cos(10), = 5*[16*sec(10)-tan(10)]/[32*x]. Now substituting this and the second equation into the second identity gives a single equation in x: 25*[16*sec(10)-tan(10)]^2/[1024*x^2] + 25/[1024*x^2] = 1, 25*[16*sec(10)-tan(10)]^2/1024 + 25/1024 = x^2, 5*sqrt([16*sec(10)-tan(10)]^2 + 25)/32 = x. If I did my calculations correctly, the solutions are approximately x = 2.515872 y = 3.560696 degrees. Then angle <BAC = 86.439314 degrees. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/