Let \(ABC\) be a triangle such that \(AB=AC\). Let \(M\) be a point on \(AB\) such that \(MA=MC\) and let \(N\) be a point on \(AC\) such that \(CN=CB\) and \(\angle BAC:\angle NBA=2:3\). Calculate the size of \(\angle NMC\).

You've found \(\angle BAC = 20^{\circ}\), but the problem is asking for \(\angle NMC\). Actually, after your solution this problem becomes literally equivalent to the one presented in Moscow 1952 and even has a name Langley's Adventitious Angles (an animated explanation here).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestJust do the angle chasing. Very easy.

Let \(B\hat{A}C = x\).

\(\displaystyle \overline{AB} = \overline{AC}; A\hat{B}C = A\hat{C}B = 90 - \frac{x}{2}\)

\(\displaystyle \overline{MA} = \overline{MC}; M\hat{C}A = x\)

\(\displaystyle \overline{CN} = \overline{CB}; C\hat{N}B = C\hat{B}N = 45 + \frac{x}{4}\)

\(\displaystyle \angle ABC; N\hat{B}A = 45 - \frac{3x}{4}\)

\(\angle BAC : \angle NBA = 2 : 3\)

\(\displaystyle 3x = 2(45 - \frac{3x}{4})\)

\(x = 20\) ~~~

Log in to reply

You've found \(\angle BAC = 20^{\circ}\), but the problem is asking for \(\angle NMC\). Actually, after your solution this problem becomes literally equivalent to the one presented in Moscow 1952 and even has a name Langley's Adventitious Angles (an animated explanation here).

Log in to reply

Please provide a full solution :D

Log in to reply