=how to solve segments cut by angle bisector

Question

given that line jk bisects angle j in triangle hjk find the indicated lenghts

hj=12 JK=15 HK=18, find KL

Answer 1

I think the bisector is JL not JK, since JK is a side of the triangle.

Cosine Rule:

18²=12²+15²-360cosHĴK, HĴK=arccos(⅛). sinHĴK=⅜√7.

HĴK=82.8192° approx. HĴL=HĴK/2=41.4096°.

Sine Rule:

sinJĤK/15=sinHĴK/18, sinJĤK=⅚sinHĴK=(5/16)√7. JĤK=55.7711° approx.

cosJĤK=√(1-175/256)=9/16.

JL̂K=HĴL+JĤK=97.1808° approx.

KL/sin41.4096°=15/sin97.1808°, KL=10.

We can also work with trig identities. For any angles A, B:

cosA=1-2sin²(A/2), sin(A/2)=√{(1-cosA)/2}. When cosA=cosHĴK=⅛, sin(A/2)=√7/4. So sinHĴL=sin(HĴK/2)=√7/4; cosHĴL=¾.

sin(A+B)+sinAcosB+cosAsinB;

sin(JL̂K)=sin(HĴL+JĤK)=(√7/4)(9/16)+(¾)(5/16)√7=9√7/64+15√7/64=⅜√7.

KL=(15√7/4)/(⅜√7)=10.