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Question Number 65884 by Sayantan chakraborty last updated on 05/Aug/19

Commented by Sayantan chakraborty last updated on 05/Aug/19

$$\mathrm{answer}\mathrm{is}\mathrm{unknown}\mathrm{to}\mathrm{me}$$

Commented by Tony Lin last updated on 06/Aug/19

$$let\mathrm{\angle }BAD=\mathrm{\angle }BCD=\theta ,\mathrm{\angle }BDA=\mathrm{\varnothing }$$$$\frac{5}{sin\mathrm{\varnothing }}=\frac{6sin\theta }{sin\theta }=6$$$$\Rightarrow sin\mathrm{\varnothing }=\frac{5}{6},cos\mathrm{\varnothing }=-\frac{\sqrt{11}}{6}$$$$2({DM}^2+{AM}^2)={DC}^2+{AD}^2$$$${DC}^2={\left(6cos\theta \right)}^2=36{cos}^2\theta$$$$\frac{AD}{sin(180°-\theta -\mathrm{\varnothing })}=\frac{6sin\theta }{sin\theta }=6$$$$AD=6sin(\theta +\mathrm{\varnothing })=5cos\theta -\sqrt{11}sin\theta$$$${AD}^2=25{cos}^2\theta -10\sqrt{11}sin\theta cos\theta +11{sin}^2\theta$$$${AC}^2=5^2+6^2-2\times 5\times 6cos(270°-2\theta -\mathrm{\varnothing })$$$$=61+60sin(2\theta +\mathrm{\varnothing })$$$$=61+60[2sin\theta cos\theta cos\mathrm{\varnothing }+(1-2{sin}^2\theta )sin\mathrm{\varnothing }]$$$$=61-20\sqrt{11}sin\theta cos\theta +50-100{sin}^2\theta$$$${AM}^2=\frac{61}{4}-5\sqrt{11}sin\theta cos\theta +\frac{25}{2}-25{sin}^2\theta$$$$2({DM}^2+{AM}^2)={DC}^2+{AD}^2$$$$2{DM}^2+\frac{61}{2}-10\sqrt{11}sin\theta cos\theta +25-50{sin}^2\theta$$$$=36{cos}^2\theta +25{cos}^2\theta -10\sqrt{11}sin\theta cos\theta +11{sin}^2\theta$$$$2{DM}^2+\frac{61}{2}+25=61({sin}^2\theta +{cos}^2\theta )$$$${DM}^2=\frac{11}{4}\Rightarrow DM=\frac{\sqrt{11}}{2}=\frac{\sqrt{a}}{2}$$$$\Rightarrow a=11$$

Commented by Sayantan chakraborty last updated on 17/Aug/19

$$\mathbf{sir}\mathbf{please}\mathbf{solve}\mathbf{question}\mathbf{number}65886$$

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