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Question Number 191102 by Mingma last updated on 18/Apr/23

Answered by mahdipoor last updated on 18/Apr/23

$$AB=AC=a$$$$AM=BC=b=\sqrt{a^2+a^2-2a^{2}cos20}=2asin10$$$${BM}^2={AM}^2+{AB}^2-2.AB.AM.cos20\Rightarrow$$$${BM}^2=4a^2{sin}^{2}10+a^2-2.a.2asin10.cos20$$$$\Rightarrow BM=a\sqrt{4{sin}^{2}10+1-4sin10cos20}$$$$\frac{BM}{sin20}=\frac{AB}{sin(180-\theta )}\Rightarrow sin\theta =\frac{AB}{BM}sin20=$$$$\frac{2sin10sin20}{\sqrt{4{sin}^{2}10+1-4.sin10.cos20}}$$$$\Rightarrow \theta =10$$$$$$

Commented by Mingma last updated on 18/Apr/23

I think the answer should be 30°

Commented by a.lgnaoui last updated on 18/Apr/23

$$\theta =\measuredangle \mathrm{MAB}+\measuredangle \mathrm{MBA}=20+\measuredangle \mathrm{MBA}$$$$\Rightarrow \theta >20.$$

Answered by a.lgnaoui last updated on 18/Apr/23

$$\bullet 1)△\mathrm{ABM}\frac{\mathrm{BM}}{\sin 20}=\frac{\mathrm{AM}}{\sin \mathrm{B}1}=\frac{\mathrm{AB}}{\sin (\pi -\theta )}$$$$\mathrm{AMsin}20=\mathrm{BMsin}\mathrm{B}1\left(1\right)$$$$$$$$\mathrm{ABsin}20=\mathrm{BMsin}\theta$$$$\mathrm{BM}=\frac{\mathrm{ABsin}20}{\sin \theta }\left(2\right)$$$$\bullet 2)△\mathrm{BMC}\frac{\mathrm{BM}}{\sin \mathrm{C}}=\frac{\mathrm{BC}}{\sin \theta }=\frac{\mathrm{MC}}{\sin (\mathrm{B}-\mathrm{B}1)}$$$$\mathrm{BMsin}\theta =\mathrm{BCsin}\mathrm{C}\left(3\right)$$$$\theta =20+\mathrm{B}1\Rightarrow \mathrm{B}1=\theta -20$$$$\left(1\right)\Rightarrow \mathrm{BM}=\frac{\mathrm{AMsin}20}{\sin \mathrm{B}1}=\frac{\mathrm{AMsin}20}{\sin (\theta -20)}$$$$\left(3\right)\Rightarrow \mathrm{BM}=\frac{\mathrm{BCsin}\mathrm{C}}{\sin \theta }$$$$\Rightarrow \frac{\mathrm{AMsin}20}{\sin (\theta -20)}=\frac{\mathrm{BCsin}\mathrm{C}}{\sin \theta }$$$$\mathrm{AM}=\mathrm{BC}\Rightarrow \frac{\sin 20}{\sin (\mathit{\theta }-20)}=\frac{\sin \mathbf{C}}{\sin \mathit{\theta }}\left(4\right)$$$$\bullet 3)△\mathrm{ABC}\mathrm{AC}=\mathrm{AB}$$$$\Rightarrow \measuredangle \mathrm{B}=\measuredangle \mathrm{C};2\measuredangle \mathrm{C}+20=180$$$$\Rightarrow \measuredangle \mathrm{C}=\frac{180-20}{2}=80$$$$\left(4\right)\Rightarrow \frac{\sin 20}{\sin (\theta -20)}=\frac{\sin 80}{\sin \theta }$$$$\sin 20\sin \theta =\sin 80\sin (\theta -20)$$$$\mathrm{E}=\sin 20\sin \theta -\sin \theta \cos 20\sin 80+\cos \theta \sin 20\sin 80=0$$$$\mathrm{E}\times \frac{1}{\cos 20}$$$$\sin \theta (\tan 20-\sin 80)+\cos \theta \tan 20\sin 80=0$$$$$$$$\Rightarrow \tan \mathit{\theta }=\frac{\mathbf{tan}20\mathbf{sin}80}{\mathbf{sin}80-\mathbf{tan}20}$$$$\mathbf{tan}\mathit{\theta }=0,57735026..$$$$$$$$\mathit{\theta }=30$$$$$$

Commented by Mingma last updated on 18/Apr/23

Excellent!

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