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Question Number 212552 by mnjuly1970 last updated on 17/Oct/24

$$$$$$\stackrel{\mathrm{Q}:}{}\mathrm{In}\mathrm{A}\stackrel{\mathrm{\Delta }}{\mathrm{B}C}:cos\left(\mathrm{A}\right)+cos\left(\mathrm{B}\right)+2cos\left(\mathrm{C}\right)=2$$$$$$$$\mathrm{show}\mathrm{that}:a+b=2c◼$$$$$$

Answered by Ghisom last updated on 17/Oct/24

$$C=\pi -(A+B)\Rightarrow$$$$\cos A+\cos B-\cos (A+B)=2$$$$c=\frac{a+b}{2}\Rightarrow$$$$a^2+b^2+2ab\cos C=c^2\mathrm{etc}.\mathrm{lead}\mathrm{to}$$$$\left(1\right)\cos A=\frac{5b-3a}{4b}\Rightarrow$$$$\sin A=\frac{\sqrt{-3(a-3b)(3a-b)}}{4b}$$$$\left(2\right)\cos B=\frac{5a-3b}{4a}\Rightarrow$$$$\sin B=\frac{\sqrt{-3(a-3b)(3a-b)}}{4a}$$$$$$$$\left(3\right)\cos A\cos B-\sin A\sin B=-\frac{3a^2-2ab+3b^2}{8ab}$$$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$$$\mathrm{inserting}\mathrm{from}\mathrm{above}$$$$\Rightarrow \mathrm{true}$$

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