Question #11a8a

2 Answers
Apr 18, 2017

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˚#

Apr 18, 2017

#32°#

Explanation:

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°.