of Cosines?
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By the Law of Cosines, if a, b, and c are the sides of a triangle and A, B, and C are opposite to a, b, and c, respectively, we have:
(a) a^2 = b^2 + c^2 - 2bc*cos(A)
(b) b^2 = a^2 + c^2 - 2ac*cos(B)
(c) c^2 = a^2 + b^2 - 2ab*cos(C).
You are given a, b, and c. Use algebra to solve for cos(A), cos(B), and cos(C) and take the arccosine of both sides.
If we take a = 10, b = 11, and c = 19, we have:
(1) Angle A:
10^2 = 11^2 + 19^2 - 2(11)(19)cos(A), by substituting a, b, and c in (a)
==> cos(A) = (10^2 - 11^2 - 19^2)/[-2(11)(19)] = 191/209, by solving for cos(A)
==> A = arccos(191/209) ≈ 24.0°.
(2) Angle B:
11^2 = 10^2 + 19^2 - 2(10)(19)cos(B), by substituting a, b, and c in (b)
==> cos(B) = (11^2 - 10^2 - 19^2)/[-2(10)(19)] = 17/19, by solving for cos(B)
==> B = arccos(17/19) ≈ 26.5°.
(3) Angle C:
Using the fact that A + B + C = 180° ==> C = 180° - A - B:
C = 180° - arccos(191/209) - arccos(17/19) ≈ 129.5°.
I hope this helps!
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Are you serious? What do you think arccos(17/19) and arccos(191/209) are?
Oh, and I do have to say this: you have the audacity to call my answers inexact (actually, I have my answers in terms of fraction and arccosine, so they are exact), but, yet, you round -140/220, 340/380, and 382/418. Seriously?
(a) a^2 = b^2 + c^2 - 2bc*cos(A)
(b) b^2 = a^2 + c^2 - 2ac*cos(B)
(c) c^2 = a^2 + b^2 - 2ab*cos(C).
You are given a, b, and c. Use algebra to solve for cos(A), cos(B), and cos(C) and take the arccosine of both sides.
If we take a = 10, b = 11, and c = 19, we have:
(1) Angle A:
10^2 = 11^2 + 19^2 - 2(11)(19)cos(A), by substituting a, b, and c in (a)
==> cos(A) = (10^2 - 11^2 - 19^2)/[-2(11)(19)] = 191/209, by solving for cos(A)
==> A = arccos(191/209) ≈ 24.0°.
(2) Angle B:
11^2 = 10^2 + 19^2 - 2(10)(19)cos(B), by substituting a, b, and c in (b)
==> cos(B) = (11^2 - 10^2 - 19^2)/[-2(10)(19)] = 17/19, by solving for cos(B)
==> B = arccos(17/19) ≈ 26.5°.
(3) Angle C:
Using the fact that A + B + C = 180° ==> C = 180° - A - B:
C = 180° - arccos(191/209) - arccos(17/19) ≈ 129.5°.
I hope this helps!
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Are you serious? What do you think arccos(17/19) and arccos(191/209) are?
Oh, and I do have to say this: you have the audacity to call my answers inexact (actually, I have my answers in terms of fraction and arccosine, so they are exact), but, yet, you round -140/220, 340/380, and 382/418. Seriously?
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Do you even know what arccos(x) even is? It is the arccosine or inverse cosine function. Get a clue, please.
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