Question #11a8a

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Answer 1 · 2017-04-18T20:42:49

The Law of Cosines states that, for a triangle with sides #a,b,# and #c# with the angle #C# opposite side #c#, that:

#c^2=a^2+b^2-2abcos(C)#

Angle #C# is opposite side #c=bar(AB)=8#. We also see that side #a=bar(BC)=12# and #b=bar(CA)=15#. Plugging these into the Law gives:

#8^2=12^2+15^2-2(12)(15)cos(C)#

#64=144+225-360cos(C)#

#-305=-360cos(C)#

#cos(C)=305/360=61/72#

Solving for #C#:

#C=cos^-1(61/72)=32˚#

Answer 2 · 2017-04-18T20:54:45

#32°#

From the figure #a=12cm, b=15cm, c=8cm#

Now just apply the formula,

#cosC= (a^2 +b^2-c^2)/(2ab) #

# cosC=(305/360)#

# cosC=(61/72)#

#61/72# is close to #sqrt3/2#

So #C=cos^-1(61/72) approx 30^@ #

You get angle C as 32°.