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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 

Although there was no illustration in the original question, one is 
roughly drawn below:

               /^\<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 

         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, 

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/   
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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