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Question Number 29723 by ajfour last updated on 11/Feb/18

Commented by ajfour last updated on 11/Feb/18

Express c in terms of a, b .

$${Express}\:\boldsymbol{{c}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:. \\ $$

Answered by mrW2 last updated on 11/Feb/18

Commented by ajfour last updated on 12/Feb/18

Thank you Sir.

$${Thank}\:{you}\:{Sir}. \\ $$

Commented by mrW2 last updated on 11/Feb/18

cos α=(b/a) β=π−α CB=(√(a^2 −b^2 )) BD=b−(√(a^2 −b^2 )) c^2 =a^2 +(b−(√(a^2 −b^2 )))^2 −2a(b−(√(a^2 −b^2 )))cos β c^2 =a^2 +(b−(√(a^2 −b^2 )))^2 +2a(b−(√(a^2 −b^2 )))cos α c^2 =a^2 +(b−(√(a^2 −b^2 )))^2 +2a(b−(√(a^2 −b^2 )))(b/a) c^2 =a^2 +(b−(√(a^2 −b^2 )))^2 +2b(b−(√(a^2 −b^2 ))) c^2 =a^2 +b^2 +a^2 −b^2 −2b(√(a^2 −b^2 ))+2b^2 −2b(√(a^2 −b^2 )) c^2 =2(a^2 +b^2 −2b(√(a^2 −b^2 ))) ⇒c=(√(2(a^2 +b^2 −2b(√(a^2 −b^2 )))))

$$\mathrm{cos}\:\alpha=\frac{{b}}{{a}} \\ $$$$\beta=\pi−\alpha \\ $$$${CB}=\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${BD}={b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}{a}\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\mathrm{cos}\:\beta \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}{a}\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\mathrm{cos}\:\alpha \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}{a}\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)\frac{{b}}{{a}} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}{b}\left({b}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right) \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${c}^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow{c}=\sqrt{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)} \\ $$

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