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Question Number 113847 by deepraj123 last updated on 15/Sep/20

$$\mathrm{The}\mathrm{two}\mathrm{adjacent}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{cyclic}$$$$\mathrm{quadrilateral}\mathrm{are}2\mathrm{and}5\mathrm{and}\mathrm{the}$$$$\mathrm{angle}\mathrm{between}\mathrm{them}\mathrm{is}60°.\mathrm{If}\mathrm{the}\mathrm{third}$$$$\mathrm{side}\mathrm{is}3,\mathrm{the}\mathrm{remaining}\mathrm{fourth}\mathrm{side}\mathrm{is}$$

Answered by 1549442205PVT last updated on 16/Sep/20

Commented by 1549442205PVT last updated on 16/Sep/20

$$\mathrm{i})\mathrm{Case}\mathrm{CD}=3$$$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}$$$$\mathrm{AB}=2,\mathrm{BC}=5,\widehat{\mathrm{ABC}}=60°,\mathrm{CD}=3\mathrm{and}$$$$\mathrm{cosine}\mathrm{theorem}\mathrm{we}\mathrm{have}$$$${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2-2\mathrm{AB}.\mathrm{BC}.\cos 60°$$$$\iff {\mathrm{AC}}^2=2^2+5^2-2.2.5.\frac{1}{2}=19.\mathrm{On}\mathrm{the}$$$$\mathrm{other}\mathrm{hand},\widehat{\mathrm{ADC}}=180°-\widehat{\mathrm{ABC}}=120°$$$$\left(\mathrm{since}\mathrm{quadrilateral}\mathrm{ABCD}\mathrm{inscribed}\right)$$$$\mathrm{Apply}\mathrm{cosine}\mathrm{theorem}\mathrm{for}\mathrm{\Delta }\mathrm{ACD}\mathrm{with}$$$$\mathrm{AC}=\sqrt{19},\mathrm{AD}=3,\widehat{\mathrm{ADC}}=120°\mathrm{we}\mathrm{get}$$$${\mathrm{AC}}^2={\mathrm{CD}}^2+{\mathrm{AD}}^2-2\mathrm{AD}.\mathrm{CDcos}120°$$$$\iff 19=3^2+{\mathrm{AD}}^2-2.3.\mathrm{AD}.(-\frac{1}{2})$$$$\iff {\mathrm{AD}}^2+3\mathrm{AD}-10=0.\mathrm{Solve}\mathrm{this}\mathrm{equation}$$$$\mathrm{\Delta }=9+4.10=7^2\Rightarrow \mathrm{AD}=(-3+7)/2=2$$$$\mathrm{Thus},\mathrm{fourth}\mathrm{side}\mathrm{equal}\mathrm{to}2\left(\mathrm{cm}\right)$$$$\mathrm{ii})\mathrm{Case}\mathrm{AD}=3\mathrm{we}\mathrm{do}\mathrm{similarly}\mathrm{also}$$$$\mathrm{getting}\mathrm{CD}=2$$

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