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 June 1st, 2017, 11:28 AM #1 Newbie   Joined: Nov 2016 From: USA Posts: 26 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 non-right 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: 2,655 Thanks: 1327 the method you used is fine ... you could have used the law of cosines twice a = 16, b = 11, C = 103 deg $c = \sqrt{a^2+b^2-2ab\cos{C}}$ then $A = \cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right) = \cos^{-1}\left(\dfrac{b-a\cos{C}}{c}\right)$ Thanks from Seventy7 Last edited by skeeter; June 1st, 2017 at 11:56 AM.
 June 1st, 2017, 12:12 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,137 Thanks: 1415 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. Thanks from Seventy7

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