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June 1st, 2017, 10:28 AM   #1
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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 10:31 AM.
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June 1st, 2017, 10:50 AM   #2
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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 $A = \cos^{-1}\left(\dfrac{b^2+c^2-a^2}{2bc}\right) = \cos^{-1}\left(\dfrac{b-a\cos{C}}{c}\right)$
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Last edited by skeeter; June 1st, 2017 at 10:56 AM.
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June 1st, 2017, 11:12 AM   #3
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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.
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