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Question Number 70277 by mr W last updated on 02/Oct/19

Answered by mind is power last updated on 02/Oct/19

$$\frac{a}{sin\left(\theta \right)}=\frac{2a}{Sin(\pi -3\theta )}=\frac{2a}{sin\left(3\theta \right)}$$$$\Rightarrow sin\left(3\theta \right)=2sin\left(\theta \right)$$$$sin\left(3\theta \right)=-4{sin}^3\left(\theta \right)+3sin\left(\theta \right)$$$$\Rightarrow -4{sin}^3\left(\theta \right)+sin\left(\theta \right)=0$$$$\Rightarrow sin\left(\theta \right)(1-4{sin}^2\left(\theta \right))=0\Rightarrow 1-4{sin}^2\left(\theta \right)=0$$$$\Rightarrow (1-2sin\left(\theta \right))(1+2sin\theta ))=0$$$$sin\left(\theta \right)=\frac{1}{2}\Rightarrow \theta ={30}^{°}$$

Commented by mr W last updated on 02/Oct/19

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Commented by mr W last updated on 02/Oct/19

$$canwesolvewithouttrigonometry?$$

Commented by mind is power last updated on 03/Oct/19

$$iwillTrysir$$

Answered by som(math1967) last updated on 03/Oct/19

$$Constructbisectorof\mathrm{\angle }B\left(2\alpha \right)AD$$$$\therefore \frac{AD}{DC}=\frac{a}{2a}=\frac{1}{2}\therefore AD=\frac{1}{3}AC$$$$again△ABC\sim △ABD$$$$\therefore \frac{AB}{AD}=\frac{AC}{AB}$$$$\therefore {AB}^2=AD\times AC=\frac{1}{3}{AC}^2$$$${AC}^2=3{AB}^2=3a^2$$$${AB}^2+{AC}^2=a^2+3a^2=4a^2={BC}^2$$$$\therefore \mathrm{\angle }A=90°\therefore \alpha +2\alpha =90$$$$\alpha =30$$

Commented by som(math1967) last updated on 03/Oct/19

Commented by som(math1967) last updated on 03/Oct/19

$$mrWsirIusegeometry$$

Commented by mr W last updated on 03/Oct/19

$$nicesolution,thankssir!$$

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