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Question Number 181812 by HeferH last updated on 01/Dec/22

Answered by mr W last updated on 01/Dec/22

$$\sin 30°=\frac{\sin (2\alpha +30°)}{2\cos \alpha }$$$$\sin (2\alpha +30°)=\cos \alpha =\sin (90°-\alpha )$$$$2\alpha +30°=90°-\alpha \Rightarrow \alpha =20°$$$$or$$$$2\alpha +30°=90°+\alpha \Rightarrow \alpha =60°$$

Commented by mr W last updated on 01/Dec/22

Commented by mr W last updated on 01/Dec/22

Answered by som(math1967) last updated on 01/Dec/22

$$letAD=DC=x,AC=BD=y$$$$from△ADC\frac{x}{sin\alpha }=\frac{y}{sin(180-2\alpha )}$$$$\Rightarrow x=\frac{ysin\alpha }{sin2\alpha }=\frac{y}{2cos\alpha }[sin2\alpha =2sin\alpha cos\alpha ]$$$$from△ABD$$$$\frac{x}{sin30}=\frac{y}{sin(180-30-2\alpha )}$$$$\Rightarrow \frac{y}{2sin30cos\alpha }=\frac{y}{sin(150-2\alpha )}$$$$\Rightarrow \frac{1}{cos\alpha }=\frac{1}{sin(150-2\alpha )}$$$$\Rightarrow sin(90-\alpha )=sin(150-2\alpha )$$$$\Rightarrow 90-\alpha =150-2\alpha$$$$\Rightarrow 2\alpha -\alpha =150-90$$$$\therefore \alpha =60$$$$or$$$$sin(90+\alpha )=sin(150-2\alpha )$$$$90+\alpha =150-2\alpha$$$$\Rightarrow 3\alpha =60\therefore \alpha =20$$

Commented by som(math1967) last updated on 01/Dec/22

Answered by Acem last updated on 01/Dec/22

Commented by Acem last updated on 01/Dec/22

$$5\alpha =187.5ineedrecheckmysolution$$

Answered by HeferH last updated on 01/Dec/22

Commented by HeferH last updated on 01/Dec/22

$$2\alpha -60°=\alpha$$$$\alpha =60°$$$$thankyouallforanswering,Iwantedtosee$$$$othermethods:)$$

Answered by a.lgnaoui last updated on 01/Dec/22

$$\frac{\sin 30}{\mathrm{BC}}=\frac{\sin \lambda }{\mathrm{AC}}\mathrm{BC}=\frac{\mathrm{AC}\times \sin 30}{\sin \lambda }$$$$\text{Missing left or extra right}$$$$\mathrm{AC}=\mathrm{BD}\Rightarrow =$$$$\frac{\sin \alpha }{\mathrm{BC}}=\frac{\sin 2\alpha }{\mathrm{AC}}\Rightarrow \frac{\sin \alpha }{\frac{\mathrm{AC}\times \sin 30}{\sin \lambda }}=\frac{\sin 2\alpha }{\mathrm{AC}}$$$$\frac{\sin \alpha \times \sin \lambda }{\sin 30}=\sin 2\alpha$$$$\sin \lambda =\frac{\sin 30\times \sin 2\alpha }{\sin \alpha }=\cos \alpha =$$$$\lambda =\frac{\pi }{2}-\alpha \left(1\right)$$$$\theta =\pi -2\alpha$$$$△\mathrm{ABD}$$$$2\alpha +\lambda +30=180$$$$\lambda =150-2\alpha \left(2\right)$$$$\left(1\right)\mathrm{et}\left(2\right)\Rightarrow 90-\alpha =150-2\alpha$$$$\alpha =150-90$$$$Angle\alpha =60°$$$$$$

Commented by a.lgnaoui last updated on 01/Dec/22

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