SOLUTION: Building A and B are 200m and 500m from Building C respectively. the buildings are connected by straight concrete roads so that the road between building A and B makes an angle of
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Question 1109663: Building A and B are 200m and 500m from Building C respectively. the buildings are connected by straight concrete roads so that the road between building A and B makes an angle of 25º30' with the road that connects buildings A and C. How far is building A from building B? Answer by Theo(13342) (Show Source):
you multiply the number of seconds by 3600 and add it to the degrees.
you multiply the number of minutes by 60 and add it to the degrees.
therefore 25 degrees 30 minutes is equal to 25 degrees + 30/60 = 25.5 degrees.
that's your angle formed by the road between A and B, and the road between A and C.
your triangle formed is triangle ABC.
side a is the side opposite angle A.
side b is the side opposite angle B.
side c is the side opposite angle C.
i believe you will first have to find angle B and then find angle C and then use the law of cosines to find the length of side c.
to find angle B, use the law of sines.
you will get a/sin(A) = b/sin(B).
this will become 500 / sin(25.5) = 200 / sin(B)
solve for sin(B) to get sin(B) = 200 * sin(25.5) / 500.
that will get you sin(B) = .1722044387.
then you use arcsin(.1722044387) to get angle (B) = 9.916014584 degrees.
since the sum of the angles of a triangle is always equal to 180 degrees, then subtract angle (A) and angle (B) from 180 to get angle (C) = 144.5839854 degrees.
now use the law of cosines to find the length of side c.
the law of cosines says c^2 = a^2 + b^2 - 2ab*cos(C).
since angle (C) = 144.5839854 degrees and side a = 500 and side b = 200, the formula becomes:
c^2 = 500^2 + 200^2 - 2*500*200*cos(144.5839854).
this results in c^2 = 452993.1701.
solve for c to get c = sqrt(452993.1701) = 673.047673 meters in length.
