Triangles and Law of Sines
Date: 03/27/98 at 17:32:36 From: Heathe Subject: Trigonometry I am having problems with the Law of Sines. To be specific, I don't know how to tell if the information I'm given produces more than one triangle. Solving for one is simple. That I can do. But some of the problems in my book produce two triangles, and all the info they give to solve the problem is like 2 sides and an angle or 2 angles and a side. For example: Angle A = 37 degrees Side C = 40 Side A = 28 The back of my book gives two solutions for this this problem. How do you solve it?
Date: 03/28/98 at 23:02:02 From: Doctor Scott Subject: Re: Trigonometry Hi Heathe! Using the law of sines you get: 28 40 -------- = ------- sin 37 sin C or 28 sin C = 40 sin 37 40 sin 37 sin C = --------- 28 sin C = 0.859735747 To solve this equation, we use the inverse sine function. InvSin(0.859..) = 59.28 degrees. But recall that the sin of an angle and the sin of its supplement are equal. That is, sin x = sin (180-x). So, if the sin of 59.28 is 0.8597, then so is sin (180-59.28) or sin (120.7). So another possible measure for C is 120.7. We need to see if it can work. Since angle A is 37, keeping in mind that the angles of a triangle add up to 180, we can have C = 59.28 and A = 37 (and still have some angle measure left over for B) OR C = 120.7 and A = 37 and still have some angle measure left over for B. That's why there are 2 solutions in the back of your book. Now, we have to find all of the other parts of the triangle for BOTH possible angle Bs. -Doctor Scott, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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