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Question 1112146: The sides of a triangle are 50,60, and 70. Find the length of the angle bisector from the longest side to its opposite vertex.
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
Let a=60, b=50, and c=70; and, with the usual convention, let A, B, and C be the vertices opposite sides a, b, and c, respectively. Let CP be the angle bisector of angle C, with P on side AB.
The angle bisector CP divides side AB into two pieces whose lengths are in the ratio a:b. So
 and  .
From there, there are at least a couple of ways to proceed.
(1) Use Stewart's Theorem.
Let CP=t; AP=n; and BP=m. With the segments labeled that way, Stewart's Theorem says
You can do the calculation; the result is t = 42.2507, to 4 decimal places.
(2) Use some combination of the law of sines and low of cosines.
I will just outline one possible sequence of calculations; you can do the actual calculations if you need to.
(a) Use the law of cosines in triangle ABC to find that angle C is (approximately) 78.463 degrees; then the bisected angle, ACP, is 39.2315 degrees;
(b) Use the law of cosines again in triangle ABC to find that angle A is (approximately) 57.122 degrees; and
(c) Use the law of sines in triangle ACP to find the length of t is (again, of course) 42.2507.
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