In $ riangle A B C$, if $b+c: c+a: a+b=7: 8: 9$, then the smallest angle (in radians) of that triangle is

Question

In $	riangle A B C$, if $b+c: c+a: a+b=7: 8: 9$, then the smallest angle (in radians) of that triangle is

Aicharya · Accepted Answer

$\begin{aligned} & b+c=7 k, c+a=8 k, a+b=9 k \ & \Rightarrow 2(a+b+c)=24 k \Rightarrow a+b+c=12 k \ & a=5 k ; b=4 k, c=3 k \ & 	ext { Smallest angle }=C \ & \Rightarrow \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{25+16-9}{2 	imes 5 	imes 4}=\frac{32}{40}=\frac{4}{5} \ & 	herefore C=\cos ^{-1}\left(\frac{4}{5}ight)=	ext { Smallest angle. }\end{aligned}$

In $\triangle A B C$, if $b+c: c+a: a+b=7: 8: 9$, then the smallest angle (in radians) of that triangle is

Solution