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Question Number 177694 by cortano1 last updated on 08/Oct/22

Commented by Strengthenchen last updated on 08/Oct/22

as the picture,  ((200)/(sin y))=((36)/(sin 5°))→sin y=((sin 5°×50)/9)→cos y=(√(1−sin^2 y))  x^2 +36^2 −72xcos y=200^2 →x=((72cos y±(√((72cos y)^2 −4(36^2 −200^2 ))))/2)  ac^2 +cd^2 =200^2 ,ac^2 +(cd+36)^2 =x^2   cos ∠DAC=(cd^2 −(200^2 +ac^2 ))/−400ac  ∠DAC=cos^(−1) ∠DAC  can′t upload picture,what happended?

$${as}\:{the}\:{picture}, \\ $$$$\frac{\mathrm{200}}{\mathrm{sin}\:{y}}=\frac{\mathrm{36}}{{sin}\:\mathrm{5}°}\rightarrow\mathrm{sin}\:{y}=\frac{\mathrm{sin}\:\mathrm{5}°×\mathrm{50}}{\mathrm{9}}\rightarrow\mathrm{cos}\:{y}=\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {y}} \\ $$$${x}^{\mathrm{2}} +\mathrm{36}^{\mathrm{2}} −\mathrm{72}{x}\mathrm{cos}\:{y}=\mathrm{200}^{\mathrm{2}} \rightarrow{x}=\frac{\mathrm{72cos}\:{y}\pm\sqrt{\left(\mathrm{72cos}\:{y}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{36}^{\mathrm{2}} −\mathrm{200}^{\mathrm{2}} \right)}}{\mathrm{2}} \\ $$$${ac}^{\mathrm{2}} +{cd}^{\mathrm{2}} =\mathrm{200}^{\mathrm{2}} ,{ac}^{\mathrm{2}} +\left({cd}+\mathrm{36}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\angle{DAC}=\left({cd}^{\mathrm{2}} −\left(\mathrm{200}^{\mathrm{2}} +{ac}^{\mathrm{2}} \right)\right)/−\mathrm{400}{ac} \\ $$$$\angle{DAC}=\mathrm{cos}\:^{−\mathrm{1}} \angle{DAC} \\ $$$${can}'{t}\:{upload}\:{picture},{what}\:{happended}? \\ $$$$ \\ $$

Answered by mr W last updated on 08/Oct/22

((sin β)/(200))=((sin 5°)/(36))  ⇒β=sin^(−1) ((50 sin 5°)/9)  ⇒α=90°−β−5°=85°−sin^(−1) ((50 sin 5°)/9)                                   ≈56.04°

$$\frac{\mathrm{sin}\:\beta}{\mathrm{200}}=\frac{\mathrm{sin}\:\mathrm{5}°}{\mathrm{36}} \\ $$$$\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{50}\:\mathrm{sin}\:\mathrm{5}°}{\mathrm{9}} \\ $$$$\Rightarrow\alpha=\mathrm{90}°−\beta−\mathrm{5}°=\mathrm{85}°−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{50}\:\mathrm{sin}\:\mathrm{5}°}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\approx\mathrm{56}.\mathrm{04}° \\ $$

Commented by mr W last updated on 08/Oct/22

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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