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Question-67849




Question Number 67849 by mr W last updated on 01/Sep/19
Commented by MJS last updated on 01/Sep/19
is the angle in the center 90°?
$$\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{90}°? \\ $$
Commented by mr W last updated on 01/Sep/19
not given.
$${not}\:{given}. \\ $$
Commented by Prithwish sen last updated on 01/Sep/19
Commented by Prithwish sen last updated on 01/Sep/19
(1/2)absinα=3...(i)  (1/2)cdsinα=2....(ii)  (1/2)bdsinβ=1...(iii)  find  (1/2)acsinβ
$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{absin}\alpha=\mathrm{3}…\left(\mathrm{i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cdsin}\alpha=\mathrm{2}….\left(\mathrm{ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{bdsin}\beta=\mathrm{1}…\left(\mathrm{iii}\right) \\ $$$$\mathrm{find}\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{acsin}\beta \\ $$
Commented by MJS last updated on 01/Sep/19
sin β =sin α
$$\mathrm{sin}\:\beta\:=\mathrm{sin}\:\alpha \\ $$
Commented by mr W last updated on 01/Sep/19
thanks sirs!  i used the same method.
$${thanks}\:{sirs}! \\ $$$${i}\:{used}\:{the}\:{same}\:{method}. \\ $$
Commented by Prithwish sen last updated on 01/Sep/19
thank you alot.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{alot}. \\ $$
Answered by mr W last updated on 01/Sep/19
Commented by mr W last updated on 01/Sep/19
β=π−α  sin β=sin α=λ  (1/2)abλ=1    ...(i)  (1/2)bcλ=3   ...(ii)  (1/2)cdλ=x   ...(iii)  (1/2)daλ=2   ...(iv)  (i)×(iii): ((abcdλ^2 )/4)=1×x  (ii)×(iv): ((abcdλ^2 )/4)=3×2  ⇒1×x=3×2  ⇒x=6
$$\beta=\pi−\alpha \\ $$$$\mathrm{sin}\:\beta=\mathrm{sin}\:\alpha=\lambda \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ab}\lambda=\mathrm{1}\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{bc}\lambda=\mathrm{3}\:\:\:…\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{cd}\lambda={x}\:\:\:…\left({iii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{da}\lambda=\mathrm{2}\:\:\:…\left({iv}\right) \\ $$$$\left({i}\right)×\left({iii}\right):\:\frac{{abcd}\lambda^{\mathrm{2}} }{\mathrm{4}}=\mathrm{1}×{x} \\ $$$$\left({ii}\right)×\left({iv}\right):\:\frac{{abcd}\lambda^{\mathrm{2}} }{\mathrm{4}}=\mathrm{3}×\mathrm{2} \\ $$$$\Rightarrow\mathrm{1}×{x}=\mathrm{3}×\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{6} \\ $$

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