Question

If a/cos A=b/cod B=c/cosC prove that triangle ABC is equilateral triangle

Answer 1

$\huge \red{ \underline{question}}$

In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral triangle.

$\huge \red{ \underline{proof}} \implies$

Step- by -step explanation:-

$a.t.q. \: \\ \: \: \frac{a}{cos \: A} = \frac{b}{cos \: B} = \frac{c}{cos \: C}........(1) \\ \\ use \: sine \:rule \\ \\ \implies \: \frac{a}{sin \: A} = \frac{b}{sin \: B} = \frac{c}{sin \: C} \: \\ \\ therefore \: reverse \: this \\ \\ \implies \: \frac{sin \: A}{a} = \frac{sin \: B}{b} = \frac{sin \: C}{c} .......(2) \\ \\ now \: multiplying \: equals \: by \: rquals \: from \: \\ (1) \: and \: (2) \\ \\ now \: we \: have \\ \\ \implies \: \bigg( \frac{sin \: A}{a} \bigg) \bigg( \frac{a}{cos \: A} \bigg) = \bigg( \frac{sin \: B}{b} \bigg) \bigg( \frac{b}{cos \: B} \bigg) = \bigg( \frac{sin \: C}{c} \bigg) \bigg( \frac{c}{cos \: C} \bigg) \\ \\ \implies \: \bigg( \frac{sin \: A}{cos \: A} \bigg) = \bigg( \frac{sin \: B}{cos \: B} \bigg) = \bigg( \frac{sin \: C}{cos \: C} \bigg) \\ \\ \implies \: tan \: A \: = tan \: B \: = tan \: C \\ \\ \implies \: A ,\: B, \: and \: C \: are \: angle \: of \: a \: triangle \: which \: are\\ \: less \: then \: 180 \degree$

A= B = C

thus, ∆ABC is a equilateral triangle.

Hence proved.