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- Feb 14, 2012

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- Jan 25, 2013

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$ \angle A=15^o$

using MC as a mirror we got triangle A' MC as a symetric image of triangle AMC

then it is very easy to see that :

$ \angle NMC=25^o=\angle A+(65^o - 55^o)$

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- Feb 14, 2012

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Hi Albert, I want to thank you for participating in this problem but your answer doesn't match mine.$ \angle A=15^o$

using MC as a mirror we got triangle A' MC as a symetric image of triangle AMC

then it is very easy to see that :

$ \angle NMC=25^o=\angle A+(65^o - 55^o)$

Here is my solution:

First I let the angle NMC be \(\displaystyle \alpha\).

Consider the triangle NPM,

\(\displaystyle \sin \alpha=\dfrac{NP}{MN}\)

Consider the triangle NCP,

\(\displaystyle \sin 55^{\circ}=\dfrac{NP}{CN}\)

\(\displaystyle \therefore NP=CN\sin 55^{\circ}\)

Now, since NP can be expressed as the function of CN, I will figure out a way to express MN in terms of CN because that is one of the valid method to find the measure of the angle \(\displaystyle \alpha\).

Consider the triangle AMN,

\(\displaystyle \dfrac{MN}{\sin 15^{\circ}}=\dfrac{AM}{\sin (55+\alpha)^{\circ}}\implies\) \(\displaystyle MN=\dfrac{AM\sin 15^{\circ}}{\sin (55+\alpha)^{\circ}}\)

But \(\displaystyle \dfrac{AM}{\sin 55^{\circ}}=\dfrac{CM}{\sin 15^{\circ}}\) (from the triangle AMC)

and \(\displaystyle \dfrac{CM}{\sin 100^{\circ}}=\dfrac{BC}{\sin 70^{\circ}}\) (from the triangle CMB)

Thus,

\(\displaystyle AM=\dfrac{CM\sin 55^{\circ}}{\sin 15^{\circ}}=\dfrac{BC\sin 100^{\circ}}{\sin 70^{\circ}}\cdot\dfrac{\sin 55^{\circ}}{\sin 15^{\circ}}\) but \(\displaystyle \dfrac{BC}{\sin 35^{\circ}}=\dfrac{CN}{\sin 80^{\circ}}\) (from the triangle BCN)

\(\displaystyle \therefore AM=\dfrac{CN\sin 35^{\circ}}{\sin 80^{\circ}}\cdot\dfrac{\sin 100^{\circ}}{\sin 70^{\circ}}\cdot\dfrac{\sin 55^{\circ}}{\sin 15^{\circ}}\)

or

\(\displaystyle MN=\dfrac{\sin 15^{\circ}\dfrac{CN\sin 35^{\circ}}{\sin 80^{\circ}}\cdot\dfrac{\sin 100^{\circ}}{\sin 70^{\circ}}\cdot\dfrac{\sin 55^{\circ}}{\sin 15^{\circ}}}{\sin (55+\alpha)^{\circ}}\)

We have now:

\(\displaystyle \sin \alpha=\dfrac{NP}{MN}\)

\(\displaystyle \sin \alpha=\dfrac{CN\sin 55^{\circ}}{\dfrac{\sin 15^{\circ}\dfrac{CN\sin 35^{\circ}}{\sin 80^{\circ}}\cdot\dfrac{\sin 100^{\circ}}{\sin 70^{\circ}}\cdot\dfrac{\sin 55^{\circ}}{\sin 15^{\circ}}}{\sin (55+\alpha)^{\circ}}}\)

Simplifying the above yields

\(\displaystyle \sin \alpha=2\cos35^{\circ}\sin (55+\alpha)^{\circ}\)

This gives

\(\displaystyle \tan \alpha=\dfrac{2\cos35^{\circ}\sin55^{\circ}}{1-2\cos35^{\circ}\sin55^{\circ}}\)

Finally, we get \(\displaystyle \angle NMC=\alpha=87.4270^{\circ}\)

So Albert, who is right and who is wrong?

BTW, may I look at your diagram because I have a little hard time trying to figure out your solution, please?

Thanks.

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- Jan 25, 2013

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Your diagram is not correct ,please check it

note :angle MCB=$55^o$,

but your diagram angle MCB=$10^o$

do it again ,I will post my solution later

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- #5

- Feb 14, 2012

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Yes, you're right, I labeled the angle incorrectly and by redoing the problem, I found that \(\displaystyle \angle NMB=25^{\circ}\).

Your diagram is not correct ,please check it

note :angle MCB=$55^o$,

but your diagram angle MCB=$10^o$

do it again ,I will post my solution later

Your approach must be one of the smartest one because you solved it by observation but I approached it using lots of messy equations and therefore, my hat is off to you, Albert!

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- Mar 5, 2012

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Ah, but you don't know how much paper and messy equations he spent before giving a short answer.Your approach must be one of the smartest one because you solved it by observation but I approached it using lots of messy equations and therefore, my hat is off to you, Albert!

- Jan 25, 2013

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- Feb 21, 2013

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Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?

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- Mar 5, 2012

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It's the angle at point B between the sides AB and BC.Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?

To be honest, I was puzzling myself which angle was intended exactly, but looking at the other angles, I deduced that this is what was intended.

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Asking for clarification about a step taken or notation used is encouraged, as this is how we learn, and is not considered off-topic at all.Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?