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Question Number 128764 by behi83417@gmail.com last updated on 10/Jan/21

Commented by behi83417@gmail.com last updated on 10/Jan/21

$$\measuredangle \mathrm{DAB}=\measuredangle \mathrm{CBA}={30}^{\bullet }$$$$\mathbf{AB}=?$$

Answered by Olaf last updated on 10/Jan/21

$$\mathrm{Let}{\mathrm{H}}_{\mathrm{D}}={\mathrm{proj}}_{\mathrm{\perp }}{\left(\mathrm{D}\right)}_{/\mathrm{AB}}$$$$\mathrm{Let}{\mathrm{H}}_{\mathrm{C}}={\mathrm{proj}}_{\mathrm{\perp }}{\left(\mathrm{C}\right)}_{/\mathrm{AB}}$$$${\mathrm{AH}}_{\mathrm{D}}=\mathrm{ADcos}30°=7\times \frac{\sqrt{3}}{2}=\frac{7\sqrt{3}}{2}$$$${\mathrm{BH}}_{\mathrm{C}}=\mathrm{BCcos}30°=4\times \frac{\sqrt{3}}{2}=2\sqrt{3}$$$${\mathrm{DH}}_{\mathrm{D}}^2={\mathrm{AD}}^2-{\mathrm{AH}}_{\mathrm{D}}^2=49-\frac{49\times 3}{4}=\frac{49}{4}$$$$\Rightarrow {\mathrm{DH}}_{\mathrm{D}}=\frac{7}{2}$$$${\mathrm{CH}}_{\mathrm{C}}^2={\mathrm{BC}}^2-{\mathrm{BH}}_{\mathrm{C}}^2=16-4\times 3=4$$$$\Rightarrow {\mathrm{CH}}_{\mathrm{C}}=2$$$$\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{such}\widehat{\mathrm{CED}}=90°$$$$\mathrm{DE}={\mathrm{DH}}_{\mathrm{D}}-{\mathrm{CH}}_{\mathrm{C}}=\frac{7}{2}-2=\frac{3}{2}$$$${\mathrm{CE}}^2={\mathrm{CD}}^2-{\mathrm{DE}}^2=25-\frac{9}{4}=\frac{91}{4}$$$$\Rightarrow \mathrm{CE}=\frac{\sqrt{91}}{2}$$$$\mathrm{AB}={\mathrm{AH}}_{\mathrm{D}}+\mathrm{CE}+{\mathrm{BH}}_{\mathrm{C}}=\frac{7\sqrt{3}}{2}+\frac{\sqrt{91}}{2}+2\sqrt{3}$$$$\mathrm{AB}=\frac{11\sqrt{3}+\sqrt{91}}{2}$$

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