A = 28 degrees, b=15, c=23
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solve the triangle using the law of cosine?
A = 28 degrees, b=15, c=23
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a^2 = b^2 + c^2 - 2bc*cos(A)
a^2 = 15^2 + 23^2 - 2*15*23*cos(28°)
a^2 = 225 + 529 - 690*cos(28°)
a = sqrt(754 - 690*cos(28°)), note that a length can't be negative
a =~ 12.031881021990718222454580187535
b^2 = a^2 + c^2 - 2ac*cos(B)
2ac*cos(B) = a^2 - b^2 + c^2
cos(B) = (a^2 - b^2 + c^2) / (2ac)
cos(B) = (754 - 690*cos(28°) - 15^2 + 23^2) / (2*sqrt(754 - 690*cos(28°))*23)
cos(B) = (754 - 690*cos(28°) - 225 + 529) / (46*sqrt(754 - 690*cos(28°)))
cos(B) = (1058 - 690*cos(28°)) / (46*sqrt(754 - 690*cos(28°)))
B = arccos((23 - 15*cos(28°)) / sqrt(754 - 690*cos(28°)))
B =~ 35.8231°
c^2 = a^2 + b^2 - 2ab*cos(C)
2ab*cos(C) = a^2 + b^2 - c^2
cos(C) = (a^2 + b^2 - c^2) / (2ab)
cos(C) = (754 - 690*cos(28°) + 15^2 - 23^2) / (2*sqrt(754 - 690*cos(28°))*15)
cos(C) = (754 - 690*cos(28°) + 225 - 529) / (30*sqrt(754 - 690*cos(28°)))
cos(C) = (450 - 690*cos(28°)) / (30*sqrt(754 - 690*cos(28°)))
C = arccos((15 - 23*cos(28°)) / sqrt(754 - 690*cos(28°)))
C =~ 116.177°
solving for side a
a = √[b^2 + c^2 - 2bccos(A)]
a = √[15^2 + 23^2 - 2(15)(23)cos(28)]
a = 12 answer//
a^2 = b^2+c^2-2bcCosA
= 15^2 + 23^2 - 2(15)(23)cos(28deg)
= 225 + 529 - 690*0.88
= 754 - 607.2 = 146.80
a = 12.11
now we have
a = 12; b = 15; c = 23
Use the following formula to calculate angle A,B and C respectively
Angle A = cos^-1[(b^2+b^2-a^2)/2ab]
where “A” is the angle opposite side “a”
Angle B = cos^-1[(a^2+c^2-b^2)/2ab]
where “B” is the angle opposite side “b”
Angle C = cos^-1[(a^2+b^2-c^2)/2ab]
where “C” is the angle opposite side “c”
A = 28 degrees, b = 15, c = 23
a = 12.032
Angle ∠ B = 35.823°
Angle ∠ C = 116.177°
the law of cosine
a^2 = b^2 + c^2 - 2bc*cosA