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Question Number 170369 by cortano1 last updated on 22/May/22

Answered by som(math1967) last updated on 22/May/22

$$sin10=\frac{b}{2a}$$$$3sin10-4{sin}^{3}10=sin30$$$$\Rightarrow \frac{3b}{2a}-\frac{4b^3}{8a^3}=\frac{1}{2}$$$$\Rightarrow \frac{3a^{2}b-b^3}{2a^3}=\frac{1}{2}$$$$\Rightarrow 3a^{2}b-b^3=a^3$$$$\therefore {\mathit{a}}^3+{\mathit{b}}^3=3{\mathit{a}}^2\mathit{b}$$$$ar.△ABC=\frac{1}{2}\times b\times \sqrt{a^2-\frac{b^2}{4}}$$$$=\frac{1}{2}\times b\times \frac{\sqrt{4a^2-b^2}}{2}$$$$=\frac{1}{4}\times \sqrt{4a^2{b}^2-b^4}$$$$=\frac{1}{4}\times \sqrt{b(4a^{2}b-b^3)}$$$$=\frac{1}{4}\times \sqrt{b(4a^{2}b-3a^{2}b+a^3)}$$$$[\because a^3+b^3=3a^{2}b]$$$$=\frac{1}{4}\times \sqrt{{ba}^2(a+b)}$$$$=\frac{1}{4}\times a\times \sqrt{b(a+b)}$$$$\frac{AD}{sin80}=\frac{BC}{sin60}$$$$AD=\frac{b}{\sqrt{3}}\times 2\times sin80$$$$=\frac{b}{\sqrt{3}}\times 2\times \frac{\sqrt{4a^2-b^2}}{2a}$$$$=\frac{\sqrt{4a^2{b}^2-b^4}}{\sqrt{3}a}=\sqrt{\frac{b(4a^{2}b-b^3)}{3}}\times \frac{1}{a}$$$$=\sqrt{\frac{b(a^3+a^{2}b)}{3}}=a\times \frac{1}{a}\sqrt{\frac{b(a+b)}{3}}$$$$=\sqrt{\frac{b(a+b)}{3}}$$$$cos80=\frac{b}{2a}$$$$tan40=\sqrt{\frac{1-cos80}{1+cos80}}$$$$=\sqrt{\frac{2a-b}{2a+b}}$$$$$$$$=\sqrt{\frac{4a^2-b^2}{{(2a+b)}^2}}=\frac{\sqrt{4a^2-b^2}}{2a+b}$$$$\frac{2r}{b}=tan40$$$$r=\frac{1}{2}\times \frac{b\sqrt{4a^2-b^2}}{2a+b}$$$$=\frac{\sqrt{4a^2{b}^2-b^4}}{2(2a+b)}=\frac{\sqrt{b(4a^{2}b-b^3)}}{2(2a+b)}$$$$=\frac{\sqrt{{ba}^2(a+b)}}{2(2a+b)}=\frac{a\sqrt{b(a+b)}}{2(2a+b)}$$$$\frac{a}{sin80}=2R$$$$R=\frac{a}{2sin80}=\frac{a}{2\times \frac{\sqrt{4a^2-b^2}}{2a}}$$$$=\frac{a^2}{\sqrt{4a^2-b^2}}=a\times \sqrt{\frac{a^{2}b}{4a^{2}b-b^3}}$$$$=a\sqrt{\frac{a^{2}b}{a^{2}b+a^3}}=a\times \sqrt{\frac{b}{(a+b)}}$$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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