How do you find the exact value of the third side given #triangle DEF#, #d=sqrt3#, e=5, and #mangleF=pi/6#?

1 Answer
Alvin L.
Dec 13, 2017

#sqrt(13)~=3.61#

Explanation:

Since we don't know that the triangle is a right triangle, we're required to use the Cosine Law. I will also presume that #angleF# is opposite the unknown side.

The Law of Cosines states that:
#c^2=a^2+b^2-2abcos(C)#, where #angleC# is the angle opposite side #c#.

In our case, it becomes (I'll call the unknown side #x#):
#x^2=5^2+(sqrt3)^2-10sqrt3cos(pi/6)#

#x^2=25+3-10sqrt3cos(pi/6)#

#x^2=28-10sqrt3cos(pi/6)#

Next we take the square root on both sides:
#x=sqrt(28-10sqrt3cos(pi/6))#

#cos(pi/6)=sqrt3/2#

#thereforex=sqrt(28-10*3/2#

#x=sqrt(28-30/2)=sqrt(28-15)#

#x=sqrt(13)#

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