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Question Number 64459 by Tawa1 last updated on 18/Jul/19

Commented by Tony Lin last updated on 18/Jul/19

$$AR=AP=2,RC=QC=5$$$$letPB=BQ=x$$$$cos60°=\frac{{(2+x)}^2+{(5+x)}^2-7^2}{2(2+x)(5+x)}$$$$\Rightarrow (2+x)(5+x)={(2+x)}^2+{(5+x)}^2-7^2$$$$\Rightarrow x^2+7x-30=0$$$$\Rightarrow (x+10)(x-3)=0$$$$\Rightarrow x=3$$$$s=\frac{1}{2}(AB+BC+AC)$$$$=\frac{1}{2}(5+8+7)$$$$△ABC=\sqrt{s(s-AB)(s-BC)(s-AC)}$$$$=\sqrt{10\times 5\times 2\times 3}$$$$=10\sqrt{3}$$

Commented by Tawa1 last updated on 18/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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