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Question-206643

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Question Number 206643 by cortano21 last updated on 21/Apr/24
Answered by mr W last updated on 21/Apr/24
Commented by mr W last updated on 21/Apr/24
cos α=((x^2 +b^2 −a^2 )/(2xb)) cos β=((x^2 +b^2 −c^2 )/(2xb)) α+β=90° cos α=sin β ((x^2 +b^2 −a^2 )/(2xb))=(√(1−(((x^2 +b^2 −c^2 )/(2xb)))^2 )) (x^2 +b^2 −a^2 )^2 =(2xb)^2 −(x^2 +b^2 −c^2 )^2 2x^4 −2(a^2 +c^2 )x^2 +a^4 +c^4 −2(a^2 +c^2 −b^2 )b^2 =0 x^2 =((a^2 +c^2 ±(√(4(a^2 +c^2 −b^2 )b^2 −(a^2 −c^2 )^2 )))/2) ⇒x=(√((a^2 +c^2 ±(√(4(a^2 +c^2 −b^2 )b^2 −(a^2 −c^2 )^2 )))/2)) with a=3, b=4, c=5 x=(√((3^2 +5^2 ±(√(4(3^2 +5^2 −4^2 )4^2 −(3^2 −5^2 )^2 )))/2)) =(√(17±4(√(14))))≈5.654 or 1.426
$$\mathrm{cos}\:\alpha=\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{xb}} \\ $$$$\mathrm{cos}\:\beta=\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{xb}} \\ $$$$\alpha+\beta=\mathrm{90}° \\ $$$$\mathrm{cos}\:\alpha=\mathrm{sin}\:\beta \\ $$$$\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{xb}}=\sqrt{\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{xb}}\right)^{\mathrm{2}} } \\ $$$$\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{2}{xb}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}{x}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){x}^{\mathrm{2}} +{a}^{\mathrm{4}} +{c}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \pm\sqrt{\mathrm{4}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){b}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$${with}\:{a}=\mathrm{3},\:{b}=\mathrm{4},\:{c}=\mathrm{5} \\ $$$${x}=\sqrt{\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \pm\sqrt{\mathrm{4}\left(\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \right)\mathrm{4}^{\mathrm{2}} −\left(\mathrm{3}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\mathrm{2}}} \\ $$$$\:\:=\sqrt{\mathrm{17}\pm\mathrm{4}\sqrt{\mathrm{14}}}\approx\mathrm{5}.\mathrm{654}\:{or}\:\mathrm{1}.\mathrm{426} \\ $$
Commented by A5T last updated on 21/Apr/24
If x=(√(17−4(√(14)))), point P cannot be inside the square such that PC=5.
$${If}\:{x}=\sqrt{\mathrm{17}−\mathrm{4}\sqrt{\mathrm{14}}},\:{point}\:{P}\:{cannot}\:{be}\:{inside}\:{the} \\ $$$${square}\:{such}\:{that}\:{PC}=\mathrm{5}. \\ $$
Commented by mr W last updated on 21/Apr/24
one solution is for point P inside the square and the other solution is for point P outside the square.
$${one}\:{solution}\:{is}\:{for}\:{point}\:{P}\:{inside} \\ $$$${the}\:{square}\:{and}\:{the}\:{other}\:{solution}\:{is} \\ $$$${for}\:{point}\:{P}\:{outside}\:{the}\:{square}. \\ $$
Commented by A5T last updated on 21/Apr/24
Yea, just noting that, since the question is specific about P being inside the square.
$${Yea},\:{just}\:{noting}\:{that},\:{since}\:{the}\:{question}\:{is} \\ $$$${specific}\:{about}\:{P}\:{being}\:{inside}\:{the}\:{square}. \\ $$
Answered by dimentri last updated on 21/Apr/24
$$\:\:\underline{\underbrace{\:}} \cancel{ \:} \\ $$
Commented by A5T last updated on 21/Apr/24
(i)⇒ cosα=((x^2 −9)/(8x)) (ii)⇒ sinα=((x^2 +7)/(8x))⇒cosα=(√(1−(((x^2 +7)/(8x)))^2 )) (i)&(ii)⇒((x^2 −9)/(8x))=(√(1−(((x^2 +7)/(8x)))^2 )) ⇒x=(√(17+4(√(14))))≈5.654
$$\left({i}\right)\Rightarrow\:{cos}\alpha=\frac{{x}^{\mathrm{2}} −\mathrm{9}}{\mathrm{8}{x}} \\ $$$$\left({ii}\right)\Rightarrow\:{sin}\alpha=\frac{{x}^{\mathrm{2}} +\mathrm{7}}{\mathrm{8}{x}}\Rightarrow{cos}\alpha=\sqrt{\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} +\mathrm{7}}{\mathrm{8}{x}}\right)^{\mathrm{2}} } \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\frac{{x}^{\mathrm{2}} −\mathrm{9}}{\mathrm{8}{x}}=\sqrt{\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} +\mathrm{7}}{\mathrm{8}{x}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{17}+\mathrm{4}\sqrt{\mathrm{14}}}\approx\mathrm{5}.\mathrm{654} \\ $$

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