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Question Number 54970 by Tip Top last updated on 15/Feb/19
If   a cos A=b cos B  in a △ABC,   then ∠C  is
$$\mathrm{If}\:\:\:{a}\:\mathrm{cos}\:{A}={b}\:\mathrm{cos}\:{B}\:\:\mathrm{in}\:\mathrm{a}\:\bigtriangleup{ABC},\: \\ $$$$\mathrm{then}\:\angle{C}\:\:\mathrm{is} \\ $$
Answered by $@ty@m last updated on 15/Feb/19
Given  (a/b)=((cos B)/(cos A))  ...(1)  We know that  (a/b)=((sin A)/(sin B))  ...(2)  from (1) and (2)  ((cos B)/(cos A))  =((sin A)/(sin B))  sin Acos A=sin Bcos B  sin 2A=sin 2B  ⇒A=B  or 2A=π−2B  from 2A=π−2B  we get  A=(π/2)−B  ⇒A+B=(π/2)  ⇒C=(π/2)
$${Given} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{cos}\:{B}}{\mathrm{cos}\:{A}}\:\:…\left(\mathrm{1}\right) \\ $$$${We}\:{know}\:{that} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{sin}\:{A}}{\mathrm{sin}\:{B}}\:\:…\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{cos}\:{B}}{\mathrm{cos}\:{A}}\:\:=\frac{\mathrm{sin}\:{A}}{\mathrm{sin}\:{B}} \\ $$$$\mathrm{sin}\:{A}\mathrm{cos}\:{A}=\mathrm{sin}\:{B}\mathrm{cos}\:{B} \\ $$$$\mathrm{sin}\:\mathrm{2}{A}=\mathrm{sin}\:\mathrm{2}{B} \\ $$$$\Rightarrow{A}={B}\:\:{or}\:\mathrm{2}{A}=\pi−\mathrm{2}{B} \\ $$$${from}\:\mathrm{2}{A}=\pi−\mathrm{2}{B} \\ $$$${we}\:{get} \\ $$$${A}=\frac{\pi}{\mathrm{2}}−{B} \\ $$$$\Rightarrow{A}+{B}=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$

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