# Find the angle of NMC.

#### anemone

##### MHB POTW Director
Staff member
In the triangle ABC, we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, $\angle MCB=55^\circ$ and $\angle NBC=80^\circ$. Find $\angle NMC$.

#### Albert

##### Well-known member
In the triangle ABC, we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, $\angle MCB=55^\circ$ and $\angle NBC=80^\circ$. Find $\angle NMC$.
$\angle A=15^o$

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 :

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

#### anemone

##### MHB POTW Director
Staff member
$\angle A=15^o$

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 :

$\angle NMC=25^o=\angle A+(65^o - 55^o)$
Hi Albert, I want to thank you for participating in this problem but your answer doesn't match mine.

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.

Last edited by a moderator:

#### Albert

##### Well-known member
anemone:

note :angle MCB=$55^o$,
but your diagram angle MCB=$10^o$
do it again ,I will post my solution later

#### anemone

##### MHB POTW Director
Staff member
anemone:

note :angle MCB=$55^o$,
but your diagram angle MCB=$10^o$
do it again ,I will post my solution later
Yes, you're right, I labeled the angle incorrectly and by redoing the problem, I found that $$\displaystyle \angle NMB=25^{\circ}$$.

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!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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!
Ah, but you don't know how much paper and messy equations he spent before giving a short answer.

#### Albert

##### Well-known member
yes before giving a short and smart solution ,I racked my brain,and spent much time thinking

#### Petrus

##### Well-known member
Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?

#### Klaas van Aarsen

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

#### MarkFL

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