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Question Number 112209 by Aina Samuel Temidayo last updated on 06/Sep/20

$$\mathrm{A}\mathrm{triangle}\mathrm{ABC}\mathrm{has}\mathrm{the}\mathrm{following}$$$$\mathrm{properties}\mathrm{BC}=1,\mathrm{AB}=\mathrm{BC}\mathrm{and}\mathrm{that}$$$$\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{from}\mathrm{vertex}\mathrm{B}\mathrm{is}$$$$\mathrm{also}\mathrm{a}\mathrm{median}.\mathrm{Find}\mathrm{all}\mathrm{possible}$$$$\mathrm{triangle}\left(\mathrm{s}\right)\mathrm{with}\mathrm{its}/\mathrm{their}$$$$\mathrm{side}-\mathrm{lengths}\mathrm{and}\mathrm{angles}.$$

Commented by Aina Samuel Temidayo last updated on 07/Sep/20

$$\mathbf{It}\mathbf{should}\mathbf{be}\mathbf{AB}=\mathbf{AC}\mathrm{and}\mathrm{not}$$$$\mathrm{AB}=\mathrm{BC}\mathrm{please}.\mathrm{Can}\mathrm{you}\mathrm{try}\mathrm{solving}$$$$\mathrm{now}?$$

Commented by Aina Samuel Temidayo last updated on 07/Sep/20

$$\mathrm{Hello}.\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{question}\mathrm{is}$$$$\mathrm{incorrect}.$$

Commented by 1549442205PVT last updated on 07/Sep/20

$$\mathrm{Triangle}\mathrm{ABC}\mathrm{is}\mathrm{isosceles}\mathrm{at}\mathrm{B}\mathrm{then}$$$$\mathrm{the}\mathrm{bisector}\mathrm{of}\mathrm{the}\mathrm{angle}\widehat{\mathrm{ABC}}\mathrm{is}$$$$\mathrm{always}\mathrm{its}\mathrm{median},\mathrm{so}\mathrm{this}\mathrm{hypothesis}$$$${\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{necessary}.\mathrm{It}\mathrm{is}\mathrm{unnecessary}$$

Commented by Aina Samuel Temidayo last updated on 07/Sep/20

$$\mathrm{Your}\mathrm{solution}\mathrm{please}?$$

Commented by 1549442205PVT last updated on 07/Sep/20

$$\mathrm{This}\mathrm{is}\mathrm{always}\mathrm{true}\mathrm{for}\mathrm{\forall }\hat{\mathrm{B}}.\mathrm{The}$$$$\mathrm{property}\mathrm{of}\mathrm{two}\mathrm{amgles}\mathrm{having}\mathrm{their}$$$$\mathrm{sum}\mathrm{equal}\mathrm{to}90°:\mathrm{If}\alpha +\beta =90°\mathrm{then}$$$$\sin \alpha =\cos \beta ,\cos \alpha =\sin \alpha ,\tan \alpha =\cot \beta$$$$.\mathrm{On}\mathrm{the}\mathrm{other}\mathrm{words},$$$$\sin \alpha =\cos (90°-\alpha )....\cos \alpha =\sin (90°-\alpha )$$

Commented by 1549442205PVT last updated on 07/Sep/20

$$\mathrm{Ok}$$

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