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Question-217631

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Question Number 217631 by mnjuly1970 last updated on 17/Mar/25
Commented by mnjuly1970 last updated on 17/Mar/25
yes . that′s right.thanks alot sir
$$\:{yes}\:.\:{that}'{s}\:{right}.{thanks}\:{alot}\:{sir} \\ $$
Answered by maths2 last updated on 17/Mar/25
a=(1/(a′))=(1/(sin(10)));b=(1/(sin(70)))=(1/(b′)) c=(1/(sin(50)))=(1/(c′)) a^2 +b^2 +c^2 =(1/(sin^2 (10)))+(1/(sin^2 (70)))+(1/(sin^2 (50))) sin(10.3)=sin(30);sin(3.70)=sin(210)=−sin(30) sin(3.50)=sin(150)=sin(30) sin(10);sin(50);−sin(70) root of sin(3x)=(1/2)⇔−sin^3 (x)+3sin(x)(1−sin^2 (x))=(1/2) ⇔sin^3 (x)−(3/4)sin(x)+(1/8)=0 (1/(a′^2 ))+(1/(b′^2 ))+(1/(c′^2 ))=(((a′b′+b′c′+c′a′)^2 −2a′b′c′(a′+b′+c′))/((a′b′c′)^2 )) =(((−(3/4))^2 )/((−(1/8))^2 ))=36 correcte answer is C
$${a}=\frac{\mathrm{1}}{{a}'}=\frac{\mathrm{1}}{{sin}\left(\mathrm{10}\right)};{b}=\frac{\mathrm{1}}{{sin}\left(\mathrm{70}\right)}=\frac{\mathrm{1}}{{b}'} \\ $$$${c}=\frac{\mathrm{1}}{{sin}\left(\mathrm{50}\right)}=\frac{\mathrm{1}}{{c}'} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{10}\right)}+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{70}\right)}+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \left(\mathrm{50}\right)} \\ $$$${sin}\left(\mathrm{10}.\mathrm{3}\right)={sin}\left(\mathrm{30}\right);{sin}\left(\mathrm{3}.\mathrm{70}\right)={sin}\left(\mathrm{210}\right)=−{sin}\left(\mathrm{30}\right) \\ $$$${sin}\left(\mathrm{3}.\mathrm{50}\right)={sin}\left(\mathrm{150}\right)={sin}\left(\mathrm{30}\right) \\ $$$${sin}\left(\mathrm{10}\right);{sin}\left(\mathrm{50}\right);−{sin}\left(\mathrm{70}\right)\:{root}\:{of} \\ $$$${sin}\left(\mathrm{3}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow−{sin}^{\mathrm{3}} \left({x}\right)+\mathrm{3}{sin}\left({x}\right)\left(\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Leftrightarrow{sin}^{\mathrm{3}} \left({x}\right)−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}'^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}'^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}'^{\mathrm{2}} }=\frac{\left({a}'{b}'+{b}'{c}'+{c}'{a}'\right)^{\mathrm{2}} −\mathrm{2}{a}'{b}'{c}'\left({a}'+{b}'+{c}'\right)}{\left({a}'{b}'{c}'\right)^{\mathrm{2}} } \\ $$$$=\frac{\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }{\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} }=\mathrm{36}\:{correcte}\:{answer}\:{is}\:{C} \\ $$$$ \\ $$$$ \\ $$

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