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Q&A

Given two angles of a triangle, finding an angle formed by a median

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Problem: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$. Show that $\angle MBC=\angle BAC$. Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$. Show that $\angle MBC=\angle BAC.

$\def\vecl{\overrightarrow}$

Remark: One may of course try to use congruent or similar triangles. I am wondering if one can solve the problem by using vectors in $\mathbf{R}^2$: for instance, can one somehow show by manipulating vectors that $$ \frac{\vecl{BM}\cdot\vecl{BC}}{|\vecl{BM}||\vecl{BC}|}=\cos(\frac{\pi}{6})\ ? $$

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I think you mean $\frac\pi6$ instead of $\frac\pi3$. (2 comments)

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