Question-210667

Question Number 210667 by mr W last updated on 15/Aug/24

Answered by ajfour last updated on 16/Aug/24

let C be origin and A(p,q) q=(p/( (√3))) q=p−a 0−q=m(2a−p) ⇒ m=−(q/(2a−p))=−(q/(p−2q))=−(1/( (√3)−2)) tan θ_? =−(2+(√3)) θ_? =π−tan^(−1) (2+(√3))

$${let}\:{C}\:{be}\:{origin}\:\:{and}\:{A}\left({p},{q}\right) \\ $$$${q}=\frac{{p}}{\:\sqrt{\mathrm{3}}} \\ $$$${q}={p}−{a} \\ $$$$\mathrm{0}−{q}={m}\left(\mathrm{2}{a}−{p}\right) \\ $$$$\Rightarrow\:\:{m}=−\frac{{q}}{\mathrm{2}{a}−{p}}=−\frac{{q}}{{p}−\mathrm{2}{q}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta_{?} =−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$\theta_{?} =\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$

Commented by mr W last updated on 16/Aug/24

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Answered by A5T last updated on 15/Aug/24

Commented by A5T last updated on 15/Aug/24

((sin30°)/y)=((sin(150°−?))/(2x))⇒sin(150°−?)=(x/y) ((sin45°)/y)=((sin(135°−?))/x)⇒sin(135°−?)=(x/y)×sin45° ⇒((sin(135°−?))/(sin(150°−?)))=sin45°=((sin(45°+?))/(sin(30°+?))) ⇒?=105°

$$\frac{{sin}\mathrm{30}°}{{y}}=\frac{{sin}\left(\mathrm{150}°−?\right)}{\mathrm{2}{x}}\Rightarrow{sin}\left(\mathrm{150}°−?\right)=\frac{{x}}{{y}} \\ $$$$\frac{{sin}\mathrm{45}°}{{y}}=\frac{{sin}\left(\mathrm{135}°−?\right)}{{x}}\Rightarrow{sin}\left(\mathrm{135}°−?\right)=\frac{{x}}{{y}}×{sin}\mathrm{45}° \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{135}°−?\right)}{{sin}\left(\mathrm{150}°−?\right)}={sin}\mathrm{45}°=\frac{{sin}\left(\mathrm{45}°+?\right)}{{sin}\left(\mathrm{30}°+?\right)} \\ $$$$\Rightarrow?=\mathrm{105}° \\ $$

Commented by mr W last updated on 15/Aug/24

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Answered by mr W last updated on 15/Aug/24

Commented by mr W last updated on 15/Aug/24

$$?=\mathrm{60}+\mathrm{45}=\mathrm{105}° \\ $$

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