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Question Number 175571 by mnjuly1970 last updated on 02/Sep/22

Answered by behi834171 last updated on 03/Sep/22

$$\mathit{c}\mathit{o}\mathit{s}\frac{\mathit{A}}{2}=\sqrt{\frac{\mathit{p}(\mathit{p}-\mathit{a})}{\mathit{b}\mathit{c}}}\mathit{a}\mathit{n}\mathit{d}\mathit{s}\mathit{o}\mathit{o}\mathit{n}...$$$$\frac{\mathit{a}(\mathit{b}+\mathit{c})}{{\mathit{b}\mathit{c}\mathit{c}\mathit{o}\mathit{s}}^2\frac{\mathit{A}}{2}}=\frac{\mathit{a}(\mathit{b}+\mathit{c})}{\mathit{p}(\mathit{p}-\mathit{a})}\mathit{a}\mathit{n}\mathit{d}\mathit{s}\mathit{o}\mathit{o}\mathit{n}....$$$$\Rightarrow lhs=\frac{\mathit{a}(\mathit{b}+\mathit{c})}{\mathit{p}(\mathit{p}-\mathit{a})}+\frac{\mathit{b}(\mathit{c}+\mathit{a})}{\mathit{p}(\mathit{p}-\mathit{b})}+\frac{\mathit{c}(\mathit{a}+\mathit{b})}{\mathit{p}(\mathit{p}-\mathit{c})}=$$$$⩾\sqrt[3]{\frac{\mathit{a}\mathit{b}\mathit{c}(\mathit{a}+\mathit{b})(\mathit{b}+\mathit{c})(\mathit{c}+\mathit{a})}{{\mathit{p}}^2{\mathit{S}}^2}}⩾$$$$⩾\sqrt[3]{\frac{\mathit{a}\mathit{b}\mathit{c}\times 2\sqrt{\mathit{a}\mathit{b}}.2\sqrt{\mathit{a}\mathit{c}}.2\sqrt{\mathit{c}b}}{{\mathit{p}}^2{\mathit{S}}^2}}=$$$$=2\sqrt[3]{\frac{{\left(\mathit{a}\mathit{b}\mathit{c}\right)}^2}{{\mathit{p}}^2{\mathit{S}}^2}}=2\sqrt[3]{\frac{16R^2{S}^2}{{\mathit{p}}^2{\mathit{S}}^2}}=2\sqrt[3]{\frac{16R^2}{{\mathit{p}}^2}}⩾$$$$⩾2\sqrt[3]{16\times 4}=8.◼$$$$$$$$[p=\frac{a+b+c}{2}=\frac{2R\left(\mathrm{\Sigma }sinA\right)}{2}⩾R\times \frac{3\sqrt{3}}{2}\Rightarrow$$$$\Rightarrow {\mathit{p}}^2⩾\frac{27}{4}{R}^2\Rightarrow \frac{R^2}{{\mathit{p}}^2}⩽\frac{4}{27}<4]$$

Answered by mahdipoor last updated on 03/Sep/22

$$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=d\Rightarrow$$$$\frac{a(b+c)}{bc.{cos}^2\left(\frac{A}{2}\right)}=\frac{(d.sinA)(d.sinB+d.sinC)}{(d.sinB)(d.sinC)\left({cos}^2\left(\frac{A}{2}\right)\right)}=$$$$\frac{\left(2sin\left(\frac{A}{2}\right)cos\left(\frac{A}{2}\right)\right)\left(2sin\left(\frac{B+C}{2}\right)cos\left(\frac{B-C}{2}\right)\right)}{sin\left(B\right).sin\left(C\right).{cos}^2\left(\frac{A}{2}\right)}=$$$$\frac{4sin\left(\frac{180-(B+C)}{2}\right)sin\left(\frac{B+C}{2}\right)cos\left(\frac{B-C}{2}\right)}{sin\left(B\right).sin\left(C\right).cos\left(\frac{180-(B+C)}{2}\right)}=$$$$\frac{4cos\left(\frac{B+C}{2}\right)sin\left(\frac{B+C}{2}\right)cos\left(\frac{B-C}{2}\right)}{sin\left(B\right).sin\left(C\right).sin\left(\frac{B+C}{2}\right)}=$$$$\frac{4cos\left(\frac{B+C}{2}\right)cos\left(\frac{B-C}{2}\right)}{sin\left(B\right).sin\left(C\right)}=2\frac{cosB+cosC}{sinB.sinC}$$$$=2\frac{sinA.cosB+sinA.cosC}{sinA.sinB.sinC}$$$$\Rightarrow \Rightarrow$$$$\frac{a(b+c)}{bc.{cos}^2\left(\frac{A}{2}\right)}+\frac{c(b+a)}{ba.{cos}^2\left(\frac{C}{2}\right)}+\frac{b(a+c)}{ac.{cos}^2\left(\frac{B}{2}\right)}=$$$$2\left(\frac{sin(A+B)+sin(A+C)+sin(C+B)}{sinA.sinB.sinC}\right)=$$$$2\left(\frac{sin(180-C)+sin(180-B)+sin(C+pB)}{sin(180-(B+C)).sinB.sinC}\right)=$$$$2\left(\frac{sin\left(C\right)+sin\left(B\right)+sin(C+B)}{sin(B+C).sinB.sinC}\right)=2f(B,C)$$$$nowprovef(B,C)⩾4....$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 15/Sep/22

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

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