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Question Number 43641 by gunawan last updated on 13/Sep/18

$$\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{one}\mathrm{angle}\mathrm{of}120°,$$$$\mathrm{the}\mathrm{lengths}\mathrm{of}\mathrm{the}\mathrm{sides}\mathrm{form}\mathrm{an}\mathrm{AP}.$$$$\mathrm{If}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{greatest}\mathrm{side}\mathrm{is}$$$$7\mathrm{cm}.,\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}$$

Answered by $@ty@m last updated on 13/Sep/18

$$Letc>b>a$$$$c^2=a^2+b^2-2ab\cos C$$$$7^2=a^2+b^2-2ab\cos 120$$$$49=a^2+b^2+ab(\because \cos 120=-\frac{1}{2})$$$$......\left(1\right)$$$$\because a,b,careinAP$$$$\therefore 2b=a+c$$$$\Rightarrow 2b=a+7$$$$\Rightarrow a=2b-7...\left(2\right)$$$$\therefore from\left(1\right)$$$$49={(2b-7)}^2+b^2+(2b-7)b$$$$49=4b^2-28b+49+b^2+2b^2-7b$$$$7b^2-35b=0$$$$7b(b-5)=0$$$$\therefore b=5$$$$\Rightarrow a=2\times 5-7=3$$$$\therefore Areaof△=\sqrt{s(s-a)(s-b)(s-c)}$$$$=\sqrt{\frac{15}{2}(\frac{15}{2}-3)(\frac{15}{2}-5)(\frac{15}{2}-7)}$$$$=\frac{1}{4}\times \sqrt{15\times 9\times 5\times 1}$$$$=\frac{1}{4}\times 5\times 3\sqrt{3}$$$$=\frac{15}{4}\sqrt{3}sq.unit$$

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