The area of an acute-angled triangle ABC is 12. Given that #a = 5# and #b=8#, what is length #c#?

2 Answers
Oct 9, 2017

Not possible to solve for an acute triangle.

Explanation:

Oct 10, 2017

We know the area of the #DeltaABC=1/2a*b*sinC#

Inserting the given values we get

#12=1/2*5*8*sinC#
#=>sinC=12/20=3/5#

So #cosC=sqrt(1-sin^2C)=sqrt(1-(3/5)^2)=4/5#

Now

#c^2=a^2+b^2-2abcosC#

#c^2=5^2+8^2-2*5*8*4/5=5^2#

#=>c=5#

Now

#cosA=(a^2+c^2-b^2)/(2 ac)#

#=>cosB=(5^2+5^2-8^2)/(2* 5*5)=-0.28#

So #B > 90^@#

Hence triangle cannot be acute-angled.