SOLUTION: the sides of a triangular lot are 130m, 180 m, and 190 m. the lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the l

Question 770250: the sides of a triangular lot are 130m, 180 m, and 190 m. the lot is to be divided by a line bisecting the longest side and drawn from the opposite vertex. Find the length of the line.
Answer by josgarithmetic(39623) (Show Source): You can put this solution on YOUR website!
Draw this picture as described:

Triangle with 130 units length side at the left, 180 units segment extending rightward from the top endpoint of the 130 unit segment, and connect the two unconnected endpoints to form the 190 unit segment of the triangle. Draw another segment starting from where the 130 unit and the 180 unit sides meet and intersect in the midpoint of the 190 length side. This bisects the 190 unit segment into two 95 length portions.
Label the angle, alpha, where the 130 unit and 190 unit segments meet.
Label the angle, beta, where the 180 unit and 190 unit segments meet.

Now look at the figure for a few seconds.

Applying Law of Cosines, we have two equations from this figure.

and

Showing directly to cosines,
,
,

Your segment bisector has given you two component triangles, so you can use either of them, and again use Law of Cosines to find the length of the segment which bisects the 190 unit length. Remember, how you have two triangles each having a 95 unit length side.

I am picking the 130 unit alpha triangle.
m = the length from where the 130 and 180 unit sides meet to the middle point of the 190 unit length side;

Since we already found cos(alpha), then
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units ( or m)
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