# Geometry problem - IOQM

In $$\triangle ABC$$, $$AC = BC$$, $$\angle C=20^{\circ}$$, $$M$$ is on the side $$AC$$ and $$N$$ is on the side $$BC$$, such that $$\angle BAN=50^{\circ}, \angle ABM = 60^{\circ}$$. Find $$\angle NMB$$ in degrees.

It is INMO-IOQM level question and I am not able to solve anyhow, please help. I tried by angle chasing but up to no avail.

Trigonometric Solution

Let $$\angle NMB=\alpha$$

$$\dfrac{BN}{\sin\alpha}=\dfrac{MB}{\sin(160-\alpha)}$$,$$\triangle NMB$$,and, $$\dfrac{AB}{\sin40}=\dfrac{MB}{\sin 80}$$, $$\triangle AMB$$ (Sine Theorem)

Since $$AB=BN$$, then

$$\dfrac{\sin\alpha}{\sin(160-\alpha)}=\dfrac{\sin40}{\sin 80}=\dfrac{1}{2\cos40}$$

$$\sin(160-\alpha)=2 \sin\alpha \cos40=\sin(\alpha+40)+ \sin(\alpha-40)$$

$$\sin(160-\alpha)- \sin(\alpha-40)=\sin(\alpha+40)$$

$$2\cos60 \sin(100-\alpha)=\sin(\alpha+40)$$

$$\sin(100-\alpha)=\sin(\alpha+40)$$

1. Case

$$100-\alpha=\alpha+40 \implies \alpha=30$$

1. Case

$$100-\alpha+\alpha+40=180 \implies 140\not=180$$ , NA

• Please,explain how yoU got angles in sine rule Oct 13, 2020 at 2:37
• Let $\angle NMB=\alpha$,$\angle MNB=180-(\alpha+20)=160-\alpha)$,$\triangle NMB$,$\angle AMB=180-(80+60)=40$,$\triangle AMB$ Oct 13, 2020 at 6:35 Take a point P on AC between A and M such that $$\angle ABP=20$$

Hence $$BA=BP$$

$$\angle PMB= \angle PBM= 40$$

Hence $$PB=PM$$

$$\angle BAN= \angle BNA= 50$$

Hence $$BA=BN$$

So $$\triangle BPN$$ is equilateral triangle

$$PN=PM$$ and $$\angle MPN=40$$

Finally

$$\angle PMN= \angle PNM= 70$$

$$\angle NMB= 30$$