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Question Number 157045 by MathSh last updated on 18/Oct/21

Commented by Tawa11 last updated on 19/Oct/21

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

Answered by mr W last updated on 19/Oct/21

$$letAD=1$$$$\frac{b}{AD}=\frac{\sin (12+6)}{\sin 6}$$$$\Rightarrow b=\frac{\sin (12+6)}{\sin 6}$$$$\frac{AB}{AD}=\frac{\sin (114+18)}{\sin 18}$$$$\Rightarrow AB=\frac{\sin (114+18)}{\sin 18}$$$$\Rightarrow \frac{AB}{b}=\frac{\sin 132\sin 6}{{\sin }^{2}18}$$$$$$$$a=\sqrt{b^2+{AB}^2-2b\times AB\cos (114+12)}$$$$\frac{a}{b}=\sqrt{1+{\left(\frac{AB}{b}\right)}^2-2\left(\frac{AB}{b}\right)\cos 126}$$$$\frac{a}{b}=\sqrt{1+{\left(\frac{\sin 132\sin 6}{{\sin }^{2}18}\right)}^2-2\left(\frac{\sin 132\sin 6}{{\sin }^{2}18}\right)\cos 126}$$$$\frac{a}{b}=\frac{\sqrt{5}+1}{2}=\phi \left(goldenratio\right)$$

Commented by MathSh last updated on 19/Oct/21

$$\mathrm{Very}\mathrm{nice}\mathrm{solution}\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{Ser}$$

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