In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

1 Answer
yumean1119
Nov 1, 2017

About 39 degree.

Explanation:

First, calculate the edge #b# with the law of cosine.

#b^2=a^2+c^2-2ac cosB#
#=9^2+5^2-2*5*9*(-1/2)=151#
#b=sqrt(151)#

Then, calculate the angle #A# with the law of sine.
#a/(sinA) = b/(sinB)#
#asinB=bsinA#
#sinA=(asinB)/b=(9*sqrt(3)/2)/sqrt(151)#
#=(9sqrt(453))/302#
#≒0.6343#

Note that #A# must be smaller than #180-120=60# degree.
Thus, angle #A# is
#A=arcsin(0.6343)≒39.37 deg#.

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