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Question Number 112209 by Aina Samuel Temidayo last updated on 06/Sep/20
A triangle ABC has the following properties BC=1, AB=BC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.
$$\mathrm{A}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{properties}\:\mathrm{BC}=\mathrm{1},\:\mathrm{AB}=\mathrm{BC}\:\mathrm{and}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{angle}\:\mathrm{bisector}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{B}\:\mathrm{is} \\ $$$$\mathrm{also}\:\mathrm{a}\:\mathrm{median}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{possible} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{with}\:\mathrm{its}/\mathrm{their} \\ $$$$\mathrm{side}−\mathrm{lengths}\:\mathrm{and}\:\mathrm{angles}. \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
It should be AB=AC and not AB=BC please. Can you try solving now?
$$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\mathrm{and}\:\mathrm{not} \\ $$$$\mathrm{AB}=\mathrm{BC}\:\mathrm{please}.\:\mathrm{Can}\:\mathrm{you}\:\mathrm{try}\:\mathrm{solving} \\ $$$$\mathrm{now}? \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Hello. I think that question is incorrect.
$$\mathrm{Hello}.\:\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{question}\:\mathrm{is} \\ $$$$\mathrm{incorrect}. \\ $$
Commented by 1549442205PVT last updated on 07/Sep/20
Triangle ABC is isosceles at B then the bisector of the angle ABC^(�) is always its median,so this hypothesis isn′t necessary.It is unnecessary
$$\mathrm{Triangle}\:\mathrm{ABC}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{at}\:\mathrm{B}\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:\widehat {\mathrm{ABC}}\:\mathrm{is} \\ $$$$\mathrm{always}\:\mathrm{its}\:\mathrm{median},\mathrm{so}\:\mathrm{this}\:\mathrm{hypothesis} \\ $$$$\mathrm{isn}'\mathrm{t}\:\mathrm{necessary}.\mathrm{It}\:\mathrm{is}\:\mathrm{unnecessary} \\ $$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20
Your solution please?
$$\mathrm{Your}\:\mathrm{solution}\:\mathrm{please}? \\ $$
Commented by 1549442205PVT last updated on 07/Sep/20
This is always true for ∀B^(�) .The property of two amgles having their sum equal to 90°:If α+β=90° then sinα=cosβ,cosα=sinα,tanα=cotβ .On the other words, sinα=cos(90°−α)....cosα=sin(90°−α)
$$\mathrm{This}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{for}\:\forall\widehat {\mathrm{B}}.\mathrm{The} \\ $$$$\mathrm{property}\:\mathrm{of}\:\mathrm{two}\:\mathrm{amgles}\:\mathrm{having}\:\mathrm{their} \\ $$$$\mathrm{sum}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{90}°:\mathrm{If}\:\alpha+\beta=\mathrm{90}°\:\mathrm{then} \\ $$$$\mathrm{sin}\alpha=\mathrm{cos}\beta,\mathrm{cos}\alpha=\mathrm{sin}\alpha,\mathrm{tan}\alpha=\mathrm{cot}\beta \\ $$$$.\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{words}, \\ $$$$\mathrm{sin}\alpha=\mathrm{cos}\left(\mathrm{90}°−\alpha\right)….\mathrm{cos}\alpha=\mathrm{sin}\left(\mathrm{90}°−\alpha\right) \\ $$
Commented by 1549442205PVT last updated on 07/Sep/20
Ok
$$\mathrm{Ok} \\ $$

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