Question
If in a $\triangle\text{ABC},\frac{\cos\text{A}}{\text{a}}=\frac{\cos\text{B}}{\text{b}}=\frac{\cos\text{C}}{\text{c}},$ then find the measures of angles A, B, C.

Answer

$\frac{\cos\text{A}}{\text{a}}=\frac{\cos\text{B}}{\text{b}}=\frac{\cos\text{C}}{\text{c}}$
$\Rightarrow\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{abc}}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{abc}}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{abc}}$
$\Rightarrow\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{c}^2-\text{b}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\therefore\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{c}^2-\text{b}^2$
$2\text{b}^2=2\text{a}^2\Rightarrow\text{b = a}$
$\therefore\text{a}^2+\text{c}^2-\text{b}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$2\text{c}^2=2\text{b}^2\Rightarrow\text{c = b}$
$\text{b}^2+\text{c}^2-\text{a}^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\Rightarrow2\text{c}^2=2\text{a}^2\Rightarrow\text{c = a}$
Hence $\text{a = b = c}$
or, $\text{A = B = C}=60^{\circ}$

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