SOLUTION: Triangle ABC has side lengths AB = 8, BC = 10, and AC = 12. Find \dfrac{\cos A}{\cos B}.

Algebra ->  Trigonometry-basics -> SOLUTION: Triangle ABC has side lengths AB = 8, BC = 10, and AC = 12. Find \dfrac{\cos A}{\cos B}.      Log On


   

Question 1210259: Triangle ABC has side lengths AB = 8, BC = 10, and AC = 12. Find \dfrac{\cos A}{\cos B}.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52932) About Me  (Show Source):
Answer by greenestamps(13216) About Me  (Show Source):
You can put this solution on YOUR website!


This triangle is NOT a scale model of a 3-4-5 right triangle....

Use the law of cosines to find cos(A) and cos(B):

c%5E2=a%5E2%2Bb%5E2-2ab%2Acos%28C%29

Angle A...
100=64%2B144-192cos%28A%29
cos%28A%29=%28208-100%29%2F192=108%2F192=9%2F16

Angle B...
144=64%2B100-160cos%28B%29
cos%28B%29=%28164-144%29%2F160=20%2F160=1%2F8

cos%28A%29%2Fcos%28B%29=%289%2F16%29%2F%281%2F8%29=9%2F2

ANSWER: 9/2 or 4.5