Familiar TrianglesDate: 01/19/99 at 21:56:41 From: Noel Subject: Integrated math 2b I need help figuring out special right triangles. I've tried to start with a and b but I get confused. I don't know the next step. For example one problem is: Find a and b. Simplify radicals whenever possible. | . |60 . | . | . | .14 | . a | . | . |____________________. b I need help with 45-45-90 triangles and 30-60-90 triangles. If you have any available notes to show me step by step, I'll be very grateful. Thank you. Date: 01/20/99 at 12:35:48 From: Doctor Peterson Subject: Re: Integrated math 2b Hi, Noel. Thanks for a carefully explained question! The two special kinds of triangles are special because two sides are related in a simple way. For the 45-45-90 triangle, + | \ |45 \ | \ A| \ C | \ | \ |_ \ | | 45\ +-----------------+ A the two sides opposite the 45-degree angles are equal. You can get the other side from the Pythagorean Theorem: C = sqrt(A^2 + A^2) = sqrt(2A^2) = A * sqrt(2) That is, the hypotenuse is the square root of 2 times the other sides. For the 30-60-90 triangle, the important thing is that it's exactly half of an equilateral triangle: + + |\ /|\ | \ /6|0\ |30\ / | \ | \A A/ | \A B| \ / | \ | \ / | \ |_ \ / | \ | | 60\ /60 | 60\ +--------+ +--------+--------+ A/2 A That means that the side opposite the 30-degree angle is half the length of the hypotenuse. Again, you can get the length of the third side by the Pythagorean Theorem: B = sqrt(A^2 - (A/2)^2) = sqrt(A^2 - (A^2)/4) = sqrt((3/4)A^2) = A * sqrt(3)/2 So in your problem, a is half of 14, or 7 (just imagine completing the equilateral triangle by reflecting the triangle in side b if this isn't clear.) You can either use the Pythagorean Theorem directly on these numbers, or multiply 7 by the square root of 3 from the formula above to get the answer. If you prefer, just try memorizing these pictures: + | \ |45 \ | \ 1| \ sqrt(2) | \ | \ |_ \ | | 45\ +-----------------+ 1 + + |\ /|\ | \ /6|0\ |30\ / | \ | \ 2 2 / | \ 2 sqrt(3)| \ / | \ | \ / | \ |_ \ / | \ | | 60\ /60 | 60\ +--------+ +--------+--------+ 1 2 - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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