How do you find the area of triangle ABC given #A=171^circ, B=1^circ, C=8^circ, b=2#?

1 Answer
Nov 4, 2016

Assuming that I didn't make any calculator mistakes:
#color(white)("XXX")"Area"_triangle~~2.494953309# sq.units

Explanation:

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By the Law of Sines:
#color(white)("XXX")a/(sin(A))=b/(sin(B))=c/(sin(C))#

For the given values this means:
#color(white)("XXX")a=(sin(171^circ)*2)/(sin(1^circ))#

#color(white)("XXXx")~~17.92697937#

and
#color(white)("XXX")c=(sin(8^circ)*2)/(sin(1^circ)#

#color(white)("XXXx")~~15.94887232#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This gives a perimeter of
#color(white)("XXX")p=17.92697937+2+15.94887232#
#color(white)("XXXx")=35.87585168#
and a semi-perimeter of
#color(white)("XXX")s=35.87585168 div 2#
#color(white)("XXXx")=17.93792584#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Heron's Formula for the Area of a Triangle
#color(white)("XXX")"Area"_triangle=sqrt(s(s-a)(s-b)(s-c))#

#color(white)("XXXXXXX")~~2.494953309#