Question

If the sines of the angles of a triangle are in the ratios $3:5:7$ their cotangent are in the ratio

• A. $2:3:7$
• B. $33:65:-15$
• C. $65:33:-15$
• D. none of these

Answer 1

The correct option is B $33:65:-15$

Given:$\sin A:\sin B:\sin C=3:5:7$
$\therefore$sides $a=3k,b=5k,c=7k$
$\therefore \cot A=\frac{\cos A}{\sin A}=\frac{\cos A}{\frac{2△}{bc}}$
$=\frac{bc\cos A}{2△}$
$=\frac{bc\left[\frac{b^2+c^2-a^2}{2bc}\right]}{2△}$
$=\frac{b^2+c^2-a^2}{4△}$
$=\frac{{\left(5k\right)}^2+{\left(7k\right)}^2-{\left(3k\right)}^2}{4△}=\frac{65k^2}{4△}$ .........................$\left(1\right)$
$\cot B=\frac{\cos B}{\sin B}=\frac{\cos B}{\frac{2△}{ac}}=\frac{ac\cos B}{2△}$
$=\frac{c^2+a^2-b^2}{4△}$
$=\frac{{\left(7k\right)}^2+{\left(3k\right)}^2-{\left(5k\right)}^2}{4△}=\frac{33k^2}{4△}$ .........................$\left(2\right)$
$\cot C=\frac{\cos C}{\sin C}=\frac{\cos C}{\frac{2△}{ab}}$
$=\frac{ab\cos C}{2△}$
$=\frac{a^2+b^2-c^2}{4△}$
$=\frac{{\left(3k\right)}^2+{\left(5k\right)}^2-{\left(7k\right)}^2}{4△}=\frac{-15k^2}{4△}$ .........................$\left(3\right)$
$\therefore \cot A:\cot B:\cot C$ from $\left(1\right),\left(2\right)$ and $\left(3\right)$
$\cot A:\cot B:\cot C=\frac{65k^2}{4△}:\frac{33k^2}{4△}:\frac{-15k^2}{4△}$
$=65:33:-15$ (on simplification)