Solving Right Triangles
Date: 03/25/98 at 08:53:00 From: Barbara Adkison Subject: Solving right triangles How do you figure out the cos, sin, or tan in these problems? It says solve each triangle described. Round measures of sides to the nearest hundredth and measures of angles to the nearest minute. Here is what you are given: 1. A = 63 a = 9.7 B = ? b = ? C = 90 c = ? How do you know when to use tan, sin, or cos?
Date: 03/25/98 at 15:59:03 From: Doctor Rob Subject: Re: Solving right triangles When you know two of the angles, it is easy to get the third: A + B + C = 180. Now draw the triangle and label the parts you know with their values: o B /| / | /27| / | / | / | 9.7 /c a| / | / | / | /63 b 90| A o-----------o C Now if you know a leg and the opposite angle, you are in position to use a sine to find the hypotenuse and/or a tangent to find the adjacent leg: sin(A) = a/c and you know a and A, so you can find c; and tan(A) = a/b and you know a and A, so you can find b. The trick is to use an equation in which you know all but one part, and to use it to find the remaining unknown part. If you know a leg and the adjacent angle, you are in position to use a cosine to find the hypotenuse and/or a tangent to find the opposite leg: cos(A) = b/c tan(A) = a/b and you know b and A, so you can find c and a. If you know both legs, you can use the Pythagorean theorem to find c: c^2 = a^2 + b^2, and a tangent to find A: tan(A) = a/b. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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