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

1 answer · posted 2y ago by Snoopy‭ · last activity 2y ago by Tim Pederick‭

Question euclidean-geometry vector

#4: Post edited by $user avatar$ Snoopy‭ · 2022-09-13T11:38:33Z (about 2 years ago)

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> **Problem**: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a [median](https://en.wikipedia.org/wiki/Median_(geometry)) 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.](https://math.codidact.com/uploads/LhXVehyseBm3cLyinDUyL8CF)
$\def\vecl{\overrightarrow}$
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
$$
~~\frac{\vecl{BM}\cdot\vecl{BC}}{\|\vecl{BM}\|\|\vecl{BC}\|}=\cos(\frac{\pi}{3})\ ?~~
$$

> **Problem**: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a [median](https://en.wikipedia.org/wiki/Median_(geometry)) 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.](https://math.codidact.com/uploads/LhXVehyseBm3cLyinDUyL8CF)
$\def\vecl{\overrightarrow}$
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
$$
\frac{\vecl{BM}\cdot\vecl{BC}}{\|\vecl{BM}\|\|\vecl{BC}\|}=\cos(\frac{\pi}{6})\ ?
$$

#3: Post edited by $user avatar$ Snoopy‭ · 2022-09-11T21:45:53Z (about 2 years ago)

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> **Problem**: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a [median](https://en.wikipedia.org/wiki/Median_(geometry)) of $\triangle ABC$ Show that $\angle MBC=30^\circ$.
> ![Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$ Show that $\angle MBC=30^\circ$.](https://math.codidact.com/uploads/LhXVehyseBm3cLyinDUyL8CF)
$\def\vecl{\overrightarrow}$
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
$$
\frac{\vecl{BM}\cdot\vecl{BC}}{\|\vecl{BM}\|\|\vecl{BC}\|}=\cos(\frac{\pi}{3})\ ?
$$

> **Problem**: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a [median](https://en.wikipedia.org/wiki/Median_(geometry)) 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.](https://math.codidact.com/uploads/LhXVehyseBm3cLyinDUyL8CF)
$\def\vecl{\overrightarrow}$
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
$$
\frac{\vecl{BM}\cdot\vecl{BC}}{\|\vecl{BM}\|\|\vecl{BC}\|}=\cos(\frac{\pi}{3})\ ?
$$

#2: Post edited by $user avatar$ Snoopy‭ · 2022-09-11T21:44:56Z (about 2 years ago)

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euclidean-geometry

euclidean-geometry vector

#1: Initial revision by $user avatar$ Snoopy‭ · 2022-09-11T21:44:35Z (about 2 years ago)

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

> **Problem**: Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a [median](https://en.wikipedia.org/wiki/Median_(geometry)) of $\triangle ABC$ Show that $\angle MBC=30^\circ$.
> ![Suppose in $\triangle ABC$, $\angle BAC = 30^\circ$ and $\angle BCA = 15^\circ$. Suppose $BM$ is a median of $\triangle ABC$ Show that $\angle MBC=30^\circ$.](https://math.codidact.com/uploads/LhXVehyseBm3cLyinDUyL8CF)

$\def\vecl{\overrightarrow}$

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
$$
\frac{\vecl{BM}\cdot\vecl{BC}}{\|\vecl{BM}\|\|\vecl{BC}\|}=\cos(\frac{\pi}{3})\  ?
$$

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