Question

The sides of a triangle are in the ratio $7:8:9$ then

• A. Angles of triangle are in the ratio $14:11:6$
• B. Angles of triangle are in the ratio $15:11:6$
• C. Angles of triangle are in the ratio $14:12:6$
• D. Angles of triangle are in the ratio $14:16:18$

Answer 1

The correct option is A Angles of triangle are in the ratio $14:11:6$

$\Rightarrow$ Using law of cosines,

$c^2=a^2+b^2-2ab\cos C$

$b^2=a^2+c^2-2ac\cos B$

$a^2+b^2+c^2-2bc\cos A$

$\therefore \cos C=\frac{a^2+b^2-c^2}{2ab}\Rightarrow C={\cos }^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)$

$\cos B=\frac{c^2+a^2-b^2}{2ac}\Rightarrow B={\cos }^{-1}\left(\frac{c^2+a^2-b^2}{2ac}\right)$

$\cos A=\frac{b^2+c^2-b^2}{2bc}\Rightarrow A={\cos }^{-1}\left(\frac{b^2+c^2-a^2}{2bc}\right)$

Ratio of the sides $=7:8:9\Rightarrow \begin{array}{c}a=7k\\ b=8k\\ c=9k\end{array}\}k=constant$

$\therefore A={\cos }^{-1}\left(\frac{64k^2+81k^2-49k^2}{144k^2}\right)={\cos }^{-1}\left(\frac{96}{144}\right)$

$B={\cos }^{-1}\left(\frac{81k^2+49k^2-64k^2}{144k^2}\right)={\cos }^{-1}\left(\frac{66}{126}\right)$

$C={\cos }^{-1}\left(\frac{49k^2+64k^2-81k^2}{112k^2}\right)={\cos }^{-1}\left(\frac{32}{112}\right)$

$\therefore A:B:C={\cos }^{-1}\left(\frac{96}{144}\right):{\cos }^{-1}\left(\frac{66}{126}\right):{\cos }^{-1}\left(\frac{32}{112}\right)$

$=\frac{96}{144}:\frac{66}{126}:\frac{32}{112}$

$=\frac{4}{3}:\frac{22}{21}:\frac{4}{7}$

$=28:22:12$

$=14:11:6$