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Question Number 176816 by Ar Brandon last updated on 27/Sep/22

Commented by MJS_new last updated on 27/Sep/22

y=x∧5x+12y=60 ⇒ x=y=((60)/(17))

$${y}={x}\wedge\mathrm{5}{x}+\mathrm{12}{y}=\mathrm{60}\:\Rightarrow\:{x}={y}=\frac{\mathrm{60}}{\mathrm{17}} \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Oh! Thank you Sir MJS

Commented by cortano1 last updated on 27/Sep/22

D(0,5) ,C(5,5) ,B(12,0),A(0,0) AC: y=x ; BD : 5x+12y=60 Let F(((60)/(17)) ,((60)/(17))) α=α_1 +α_2 ⇒sin α=sin (α_1 +α_2 ) = sin α_1 cos α_2 +cos α_1 sin α_2 = (1/( (√2))) .((60)/(156)) +(1/( (√2))) .((144)/(156)) = (1/( (√2))) ((204)/(156)) =((17)/(13(√2)))= ((17(√2))/(26))

$$\mathrm{D}\left(\mathrm{0},\mathrm{5}\right)\:,\mathrm{C}\left(\mathrm{5},\mathrm{5}\right)\:,\mathrm{B}\left(\mathrm{12},\mathrm{0}\right),\mathrm{A}\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\:\mathrm{AC}:\:\mathrm{y}=\mathrm{x}\:;\:\mathrm{BD}\::\:\mathrm{5x}+\mathrm{12y}=\mathrm{60} \\ $$$$\:\mathrm{Let}\:\mathrm{F}\left(\frac{\mathrm{60}}{\mathrm{17}}\:,\frac{\mathrm{60}}{\mathrm{17}}\right) \\ $$$$\:\alpha=\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \Rightarrow\mathrm{sin}\:\alpha=\mathrm{sin}\:\left(\alpha_{\mathrm{1}} +\alpha_{\mathrm{2}} \right) \\ $$$$\:=\:\mathrm{sin}\:\alpha_{\mathrm{1}} \:\mathrm{cos}\:\alpha_{\mathrm{2}} +\mathrm{cos}\:\alpha_{\mathrm{1}} \:\mathrm{sin}\:\alpha_{\mathrm{2}} \\ $$$$\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\frac{\mathrm{60}}{\mathrm{156}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:.\frac{\mathrm{144}}{\mathrm{156}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\frac{\mathrm{204}}{\mathrm{156}}\:=\frac{\mathrm{17}}{\mathrm{13}\sqrt{\mathrm{2}}}=\:\frac{\mathrm{17}\sqrt{\mathrm{2}}}{\mathrm{26}} \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thanks! But how to get F(((60)/(17)), ((60)/(17)))?

$$\mathrm{Thanks}! \\ $$$$\mathrm{But}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{F}\left(\frac{\mathrm{60}}{\mathrm{17}},\:\frac{\mathrm{60}}{\mathrm{17}}\right)? \\ $$

Answered by mr W last updated on 27/Sep/22

AC=5(√2) DB=(√(5^2 +12^2 ))=13 area=((5×(5+12))/2)=((5(√2)×13×sin α)/2) ⇒sin α=((17)/(13(√2)))=((17(√2))/(26))

$${AC}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$$${DB}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{13} \\ $$$${area}=\frac{\mathrm{5}×\left(\mathrm{5}+\mathrm{12}\right)}{\mathrm{2}}=\frac{\mathrm{5}\sqrt{\mathrm{2}}×\mathrm{13}×\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{17}}{\mathrm{13}\sqrt{\mathrm{2}}}=\frac{\mathrm{17}\sqrt{\mathrm{2}}}{\mathrm{26}} \\ $$

Answered by HeferH last updated on 27/Sep/22

∠ DCA = 45° tan (∠DBA) = (5/(12)) ∠ DBA = tan^(−1) ((5/(12))) ≈ 22.7° a + 45° + 22.7° = 180° a = 112.3° sin (112.3°)≈ 0.9

$$\: \\ $$$$\:\angle\:{DCA}\:=\:\mathrm{45}° \\ $$$$\:\mathrm{tan}\:\left(\angle{DBA}\right)\:=\:\frac{\mathrm{5}}{\mathrm{12}}\: \\ $$$$\:\:\angle\:{DBA}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{12}}\right)\:\approx\:\mathrm{22}.\mathrm{7}° \\ $$$$\:{a}\:+\:\mathrm{45}°\:+\:\mathrm{22}.\mathrm{7}°\:=\:\mathrm{180}° \\ $$$$\:{a}\:=\:\mathrm{112}.\mathrm{3}° \\ $$$$\:\mathrm{sin}\:\left(\mathrm{112}.\mathrm{3}°\right)\approx\:\mathrm{0}.\mathrm{9} \\ $$$$\: \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thank you ��

Answered by som(math1967) last updated on 27/Sep/22

∠DAC=∠DCA=(π/4) ∴∠CAB=(π/2)−(π/4)=(π/4) BD=(√(5^2 +12^2 ))=13 ∴∠ABD=sin^(−1) (5/(13)) a=π−(π/4)−sin^(−1) (5/(13)) sina=sin(((3π)/4) −sin^(−1) (5/(13))) =sin((3π)/4)cos(sin^(−1) (5/(13)))−cos((3π)/4)sinsin^(−1) (5/(13)) =(1/( (√2)))×((12)/(13)) +(1/( (√2)))×(5/(13)) ★ =((17)/( 13(√2)))=((17(√2))/(26)) ★ cos(sin^(−1) (5/(13)))=cos(cos^(−1) ((12)/(13)))

$$\angle{DAC}=\angle{DCA}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore\angle{CAB}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{4}} \\ $$$$\:{BD}=\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{13} \\ $$$$\therefore\angle{ABD}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\:{a}=\pi−\frac{\pi}{\mathrm{4}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$${sina}={sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\:−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}}\right) \\ $$$$\:={sin}\frac{\mathrm{3}\pi}{\mathrm{4}}{cos}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}}\right)−{cos}\frac{\mathrm{3}\pi}{\mathrm{4}}{sin}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{12}}{\mathrm{13}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{5}}{\mathrm{13}}\:\bigstar \\ $$$$=\frac{\mathrm{17}}{\:\mathrm{13}\sqrt{\mathrm{2}}}=\frac{\mathrm{17}\sqrt{\mathrm{2}}}{\mathrm{26}} \\ $$$$\bigstar\:{cos}\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{13}}\right)={cos}\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Great! Thanks!

Commented by Tawa11 last updated on 28/Sep/22

Great sirs.

$$\mathrm{Great}\:\mathrm{sirs}. \\ $$

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