How do you use the law of cosines to find BC given a triangle ABC with Angle A=93, AC=17 and AB=28?

2 Answers
Dec 31, 2016

Answer:

There are 3 forms of The Law of Cosines so we pick the appropriate one and substitute in the values. Please see below.

Explanation:

Let's change the notation to be where angles are in uppercase and the side opposite that is the same letter but in lowercase:

#angle A# is unchanged #= 93^@#

Side AC becomes side #b = 17#

Side AB becomes side #c = 28#

Side BC becomes side #a = unknown#

Also unknown are #angle B and angle C# but they are not need for this problem.

The one of the three forms of the Law of Cosines that we need is:

#a^2 = b^2 + c^2 - 2(b)(c)cos(A)#

Substitute in the known values

#a^2 = 17^2 + 28^2 - 2(17)(28)cos(93^@)#

solve for a:

#a = sqrt(17^2 + 28^2 - 2(17)(28)cos(93^@))#

#a ~~ 33.5 #

Dec 31, 2016

Answer:

#abs(BC)~~33.51#

Explanation:

Given
#color(white)("XXX")/_A=93^@#

(I've assumed the "degree" part)

and using the standard convention lettering
#color(white)("XXX")a= "length of side opposite "/_ A=abs(BC)#
#color(white)("XXX")b= "length of side opposite "/_B = abs(AC)#
#color(white)("XXX")c= "length of side opposite "/_C = abs(AB)#

The Law of Cosines tells us that
#color(white)("XXX")a^2=b^2+c^2-2bc * cos(A)#

In this case
#color(white)("XXX")a^2=17^2+28^2-2 * 17 * 28 *cos(93^@)#

using a calculator
#color(white)("XXX")a^2~~1122.82383#
and
#color(white)("XXX")a~~33.51#