If the sides of `DeltaABC` in the ratio 4 : 5 : 6, prove that one angle is twice that of the other.

Explanation

Here greatest angle will be `angleC` and least angle will be `angleA`.
`cosC=(a^(2)+b^(2)-c^(2))/(2ab)=(4^(2)+5^(2)-6^(2))/( 2 xx 4 xx 5)`
`=(16+25-36)/(40)=5/40=1/8`
and `cosA=(b^(2)+c^(2)-a^(2))/(2bc)=5/40=1/8`
and `cosA=(b^(2)+c^(2)-a^(2))/(2bc)=(5^(2)+6^(2)-4^(2))/(2 xx 5 xx 6)`
`(25 + 36 -16)/(60)=45/60=3/4`
`therefore cos2A=2cos^(2)A-1`
`=2(3/4)-1=2 xx 9/16-1`
`=1/8=cosC`
`therefore 2A=C`. Hence Proved.