Advertisements
Advertisements
Question
In a triangle ABC, with usual notations, if \[\frac{2\cos\mathrm{A}}{\mathrm{a}}+\frac{\cos\mathrm{B}}{\mathrm{b}}+\frac{2\cos\mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ac}}\]then ∠A =
Options
\[\frac{\pi}{2}\]
\[\frac{\pi}{4}\]
\[\frac{\pi}{3}\]
\[\frac{\pi}{6}\]
Advertisements
Solution
\[\frac{\pi}{2}\]
Explanation:
\[\frac{2\cos\mathrm{A}}{\mathrm{a}}+\frac{\cos\mathrm{B}}{\mathrm{b}}+\frac{2\cos\mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ac}}\]
\[\therefore\quad\frac{2\left(\mathrm{b}^2+\mathrm{c}^2-\mathrm{a}^2\right)}{2\mathrm{abc}}+\frac{\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2}{2\mathrm{abc}}\] \[+\frac{2\left(\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2\right)}{2\mathrm{abc}}=\frac{2\mathrm{a}^2+2\mathrm{b}^2}{2\mathrm{abc}}\] ...[by cosine rule]
\[\therefore\quad2\mathrm{b}^2+2\mathrm{c}^2-2\mathrm{a}^2+\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2+2\mathrm{a}^2+2\mathrm{b}^2-2\mathrm{c}^2\] \[=2\mathrm{a}^2+2\mathrm{b}^2\]
∴ a² = b² + c²
∴ ABC is a right-angled triangle
∴ ∠A = \[\frac{\pi}{2}\]
