<ABC=? <BAC=30^o, <BDC=45^o, median (2021 All-Siberian Qualifying II 8.4)

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<ABC=? <BAC=30^o, <BDC=45^o, median (2021 All-Siberian Qualifying II 8.4)

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#1 h May 19, 2021, 5:59 PM • 3 Y

Y by Mango247, Mango247, Mango247

Given a triangle $ABC$ with an angle $BAC$ of $30$ . In this triangle let $BD$ be the median. The angle $BDC$ turned out to be $45^o$ . Find the angle $ABC$ .

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#2 h May 20, 2021, 6:44 PM

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Given $\angle BAD=30^{\circ},\angle BDC=45^{\circ},\angle BDA=135^{\circ},\angle ABD =15^{\circ}$ .
Let $\angle DBC=\beta$ and $\angle BDC=135^{\circ}-\beta$ .

Points $A(0,0),D(c,0),C(2c,0)$ and $B(\frac{\sqrt{3}}{2}b,\frac{b}{2})$ with $b=c(\sqrt{3}+1)$ .
Rule of sines in $\triangle ABC\ :\ \frac{2c}{\sin(\beta+15^{\circ}}=\frac{b}{\sin(45^{\circ}+\beta)}$ .

We calculate: $2\sin(45^{\circ}+\beta)=(\sqrt{3}+1)\sin(\beta+15^{\circ}$ .

We use $\sin 15^{\circ}=\sin(45^{\circ}-30^{\circ})$ and $\cos 15^{\circ}=\cos(45^{\circ}-30^{\circ})$
and find $\tan \beta=\frac{1}{\sqrt{3}}$ , so $\beta=30^{\circ}$ and $\angle ABC =45^{\circ}$ .

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#3 h May 20, 2021, 7:58 PM

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My solution at https://stanfulger.blogspot.com/2021/05/2021-all-siberian-qualifying-ii-form-8.html

Best regards,
sunken rock

This post has been edited 1 time. Last edited by sunken rock, May 20, 2021, 8:01 PM

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