Finding Side Lengths of a Scalene Triangle
Date: 6/2/96 at 19:31:27 From: Anonymous Subject: Finding side lengths of a scalene triangle This following question was on a very good university entrance exam in Brazil, in 1993. It states that: "Two observers on points A and B of a national park see a beginning fire on point C. Knowing that the angles CAB=45 degrees, ABC=105 degrees and that the distance between points A and B is of 15 kilometers, determine the distances between B and C, and between A and C." Although there was no illustration in the original question, one is roughly drawn below: B /^\<this angle measures 105 degrees 15 km> / \ / \ / \ / \ A /_ _ _ _ _ _ _ _ _\C this is 45 degrees this is, consequently, 30 degrees I tried to use the equation of areas: A= side*side*sin(of angle between the sides)/2 but I didn't get any results.
Date: 6/3/96 at 10:33:54 From: Doctor Pete Subject: Re: Finding side lengths of a scalene triangle The fact that angle BAC is 45 degrees and ACB is 30 degrees was very suggestive to me, so I drew the perpendicular from point B to side AC, which meets at point D. Then BD = AD = AB/sqrt(2) = 15/sqrt(2), since triangle ABD is 45-45-90 and thus isosceles. Also, triangle BCD is 30-60-90, so BC = 2BD = 30/sqrt(2), and CD = sqrt(3)*BD = 15*sqrt(3/2). Therefore the lengths we wish to find are: BC = 15*sqrt(2), AC = AD+CD = 15/sqrt(2)+15*sqrt(3/2) = 15(1+sqrt(3))/sqrt(2). Alternatively, you could use the Law of Sines, which states sin(A) sin(B) sin(C) ------ = ------ = ------ , a b c where A, B, C are angles and a, b, c are the lengths of the sides they subtend (are opposite to). So side AB is "c" in the above equation. sin(A) and sin(C) are easy to find; they are 1/sqrt(2) and 1/2, respectively. sin(B) = sin(105) = sin(45+60) = sin(45)cos(60)+cos(45)sin(60) = (1/sqrt(2))*(sqrt(3)/2)+(1/sqrt(2))*(1/2) = (1+sqrt(3))/(2*sqrt(2)). So we have 1/sqrt(2) (1+sqrt(3))/(2*sqrt(2)) 1/2 --------- = ----------------------- = --- , a b 15 so b = 15(1+sqrt(3))/sqrt(2) = AC, and a = 15*sqrt(2) = BC, which agrees with our previous results. In general, area considerations are a poor way of obtaining relations between angles and sides, because they are often very complicated and often come in a form that requires knowing the lengths of more than one side. If you know a lot of angles, a better approach is to think of the Law of Sines or the Law of Cosines (c^2 = a^2+b^2-2*a*b*cos(C)). Notice that the values of the angles were special because they allowed the first solution I gave. In general, given a side and two angles, you must use the Law of Sines to find the other lengths. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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