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Question Number 61449 by Tawa1 last updated on 02/Jun/19

Commented by Tawa1 last updated on 02/Jun/19

((a^2 + b^2 )/c^2 ) = ?

$$\frac{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} }\:\:=\:\:? \\ $$

Answered by perlman last updated on 02/Jun/19

(c/(sin(30)))=(a/(sin(140)))=(b/(sin(130))) b=((sin(130))/(sin(30)))c a=((sin(140))/(sin(30)))c ((a^2 +b^2 )/c^2 )=((sin^2 (130)+sin^2 (140))/(sin^2 (30)))=4(sin^2 (130)+sin^2 (140))=4(sin^2 (180−50)+sin^2 (90+50)) =4sin^2 (50)+4cos^2 (50)=4(sin^2 (50)+cos^2 (50))=4×1=4

$$\frac{{c}}{{sin}\left(\mathrm{30}\right)}=\frac{{a}}{{sin}\left(\mathrm{140}\right)}=\frac{{b}}{{sin}\left(\mathrm{130}\right)} \\ $$$${b}=\frac{{sin}\left(\mathrm{130}\right)}{{sin}\left(\mathrm{30}\right)}{c} \\ $$$${a}=\frac{{sin}\left(\mathrm{140}\right)}{{sin}\left(\mathrm{30}\right)}{c} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\frac{{sin}^{\mathrm{2}} \left(\mathrm{130}\right)+{sin}^{\mathrm{2}} \left(\mathrm{140}\right)}{{sin}^{\mathrm{2}} \left(\mathrm{30}\right)}=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{130}\right)+{sin}^{\mathrm{2}} \left(\mathrm{140}\right)\right)=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{180}−\mathrm{50}\right)+{sin}^{\mathrm{2}} \left(\mathrm{90}+\mathrm{50}\right)\right) \\ $$$$=\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{50}\right)+\mathrm{4}{cos}^{\mathrm{2}} \left(\mathrm{50}\right)=\mathrm{4}\left({sin}^{\mathrm{2}} \left(\mathrm{50}\right)+{cos}^{\mathrm{2}} \left(\mathrm{50}\right)\right)=\mathrm{4}×\mathrm{1}=\mathrm{4} \\ $$$$ \\ $$

Commented by Tawa1 last updated on 02/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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