Question 1112146
<br>
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.<br>
The angle bisector CP divides side AB into two pieces whose lengths are in the ratio a:b.  So
{{{AP = (5/11)*70 = 350/11}}} and {{{BP = (6/11)*70 = 420/11}}}.<br>
From there, there are at least a couple of ways to proceed.<br>
(1) Use Stewart's Theorem.<br>
Let CP=t; AP=n; and BP=m.  With the segments labeled that way, Stewart's Theorem says<br>
{{{ana+bmb=tct+mnc}}}<br>
{{{(60)(350/11)(60)+(50)(420/11)(50) = (t)(7)(t)+(420/11)(350/11)(70)}}}<br>
You can do the calculation; the result is t = 42.2507, to 4 decimal places.<br>
(2) Use some combination of the law of sines and low of cosines.<br>
I will just outline one possible sequence of calculations; you can do the actual calculations if you need to.<br>
(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.