
Geometry Geometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 1st, 2017, 11:28 AM  #1 
Member Joined: Nov 2016 From: USA Posts: 33 Thanks: 1  Did I solve correctly with Law of Sines & Cosines or is there a shortcut?
I solved a Geometry problem asking to find the measure of one specific angle of a particular scalene nonright triangle. The problem provides two side lengths and one angle. Given: angle C=103 degrees side c (opposite from angle C) is not given. side a = 16 side b = 11, find the degree measure of angle A (opposite from side a). First, I used Law of Cosines c^2= (16)^2 * (11)^2  2(16)(11)cos103 c^2= 256 + 121  (352)cos103 c^2= 377  (352)(0.22495) c^2= 377  (79.1828 c^2= 456.1828 length of side c= 21.36 Next, I used Law of Sines: (a/sin A) = (c/sin C) (16/sin A) = (21.36/sin 103) (16/sin A) = (21.36/0.9744) (16/sin A) = 21.921 sin A = (16/21.921) sin A = .7299 angle A= 46.88 degrees To verify my answer for Angle A, I solved for angle B (b/sin B) = (c/sin C) (11/sin B) = (21.36/sin 103) (11/sin B) = (21.36/0.9744) (11/sin B) = 21.9211 sin B = (11/21.9211) sin B = .5018 Angle B = 30.12 degrees Angle C + Angle A + Angle B = 103 + 46.88 + 30.12 = 180.0 degrees The directions say to round all decimal answers to the nearest whole number, so my answer for Angle A is 47 degrees. Did I correctly solve for Angle A (47 degrees)? Does this problem require the use of both Law of Cosines and Law of Sines, or is there a short cut method or formula that I could have used? (computer program not allowed) Last edited by Seventy7; June 1st, 2017 at 11:31 AM. 
June 1st, 2017, 11:50 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,692 Thanks: 1351 
the method you used is fine ... you could have used the law of cosines twice a = 16, b = 11, C = 103 deg $c = \sqrt{a^2+b^22ab\cos{C}}$ then $A = \cos^{1}\left(\dfrac{b^2+c^2a^2}{2bc}\right) = \cos^{1}\left(\dfrac{ba\cos{C}}{c}\right)$ Last edited by skeeter; June 1st, 2017 at 11:56 AM. 
June 1st, 2017, 12:12 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,433 Thanks: 1462 
You could use cot(A) = b/a csc(C)  cot(C) = (11/16)1.0263 + 0.23087 = 0.93645, so A = 46.88°, which rounds up to 47°. All the values I've calculated are approximate, but the equation I used is exact. 

Tags 
correctly, cosines, law, shortcut, sines, solve, solved 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
the law of sines and the law of cosines  Ala  Trigonometry  2  May 13th, 2017 02:10 PM 
OneSided Limits with Sines and Cosines  njuice8  Calculus  1  May 19th, 2013 10:01 PM 
Law of sines/cosines  jkh1919  Algebra  1  November 26th, 2011 02:27 PM 
Equivalence of the Law of Sines, PT, and Law of Cosines  molokach  Algebra  19  February 25th, 2011 05:02 PM 
Help: simplify this expression with just sines and cosines?  Stephanie7  Algebra  1  April 19th, 2009 02:04 PM 