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Question-186125




Question Number 186125 by Mingma last updated on 01/Feb/23
Answered by som(math1967) last updated on 01/Feb/23
let 2^(sinα) =3^(sinβ) =5^(sinθ) =k  ⇒k^(1/(sinα)) =2⇒k^(cosecα) =2  ⇒k^(cosecβ) =3  ⇒k^(cosec θ) =5  k^(cosec α+cosec β+cosec θ) =2×3×5  ⇒k^8 =30  ∴k=(30)^(1/8)   2^(sinα) +2^(sin β) +2^(sin θ)   =k+k+k  =3k=3×(30)^(1/8)
$${let}\:\mathrm{2}^{{sin}\alpha} =\mathrm{3}^{{sin}\beta} =\mathrm{5}^{{sin}\theta} ={k} \\ $$$$\Rightarrow{k}^{\frac{\mathrm{1}}{{sin}\alpha}} =\mathrm{2}\Rightarrow{k}^{{cosec}\alpha} =\mathrm{2} \\ $$$$\Rightarrow{k}^{{cosec}\beta} =\mathrm{3} \\ $$$$\Rightarrow{k}^{\mathrm{cosec}\:\theta} =\mathrm{5} \\ $$$${k}^{\mathrm{cosec}\:\alpha+\mathrm{cosec}\:\beta+\mathrm{cosec}\:\theta} =\mathrm{2}×\mathrm{3}×\mathrm{5} \\ $$$$\Rightarrow{k}^{\mathrm{8}} =\mathrm{30}\:\:\therefore{k}=\left(\mathrm{30}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \\ $$$$\mathrm{2}^{{sin}\alpha} +\mathrm{2}^{\mathrm{sin}\:\beta} +\mathrm{2}^{\mathrm{sin}\:\theta} \\ $$$$={k}+{k}+{k} \\ $$$$=\mathrm{3}{k}=\mathrm{3}×\left(\mathrm{30}\right)^{\frac{\mathrm{1}}{\mathrm{8}}} \\ $$

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