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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=(√(a^2 +a^2 −2a^2 cos20))=2asin10 BM^2 =AM^2 +AB^2 −2.AB.AM.cos20⇒ BM^2 =4a^2 sin^2 10+a^2 −2.a.2asin10.cos20 ⇒BM=a(√(4sin^2 10+1−4sin10cos20)) ((BM)/(sin20))=((AB)/(sin(180−θ)))⇒sinθ=((AB)/(BM))sin20= ((2sin10sin20)/( (√(4sin^2 10+1−4.sin10.cos20)))) ⇒θ=10

$${AB}={AC}={a} \\ $$$${AM}={BC}={b}=\sqrt{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {cos}\mathrm{20}}=\mathrm{2}{asin}\mathrm{10} \\ $$$${BM}^{\mathrm{2}} ={AM}^{\mathrm{2}} +{AB}^{\mathrm{2}} −\mathrm{2}.{AB}.{AM}.{cos}\mathrm{20}\Rightarrow \\ $$$${BM}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \mathrm{10}+{a}^{\mathrm{2}} −\mathrm{2}.{a}.\mathrm{2}{asin}\mathrm{10}.{cos}\mathrm{20} \\ $$$$\Rightarrow{BM}={a}\sqrt{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{10}+\mathrm{1}−\mathrm{4}{sin}\mathrm{10}{cos}\mathrm{20}} \\ $$$$\frac{{BM}}{{sin}\mathrm{20}}=\frac{{AB}}{{sin}\left(\mathrm{180}−\theta\right)}\Rightarrow{sin}\theta=\frac{{AB}}{{BM}}{sin}\mathrm{20}= \\ $$$$\frac{\mathrm{2}{sin}\mathrm{10}{sin}\mathrm{20}}{\:\sqrt{\mathrm{4}{sin}^{\mathrm{2}} \mathrm{10}+\mathrm{1}−\mathrm{4}.{sin}\mathrm{10}.{cos}\mathrm{20}}} \\ $$$$\Rightarrow\theta=\mathrm{10} \\ $$$$ \\ $$

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

θ=∡MAB+∡MBA=20+∡MBA ⇒θ >20.

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

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

•1)△ABM ((BM)/(sin 20))=((AM)/(sin B1))=((AB)/(sin (π−θ))) AMsin 20=BMsin B1 (1) ABsin 20=BMsin θ BM=((ABsin 20)/(sin θ)) (2) •2)△BMC ((BM)/(sin C))=((BC)/(sin θ))=((MC)/(sin (B−B1))) BMsin θ=BCsin C (3) θ=20+B1⇒B1=θ−20 (1)⇒BM=((AMsin 20)/(sin B1))=((AMsin 20)/(sin (θ−20))) (3)⇒BM=((BCsin C)/(sin θ)) ⇒ ((AMsin 20)/(sin (θ−20)))=((BCsin C)/(sin θ)) AM=BC⇒ ((sin 20)/(sin (𝛉−20)))=((sin C)/(sin 𝛉)) (4) •3)△ABC AC=AB ⇒ ∡B=∡C ; 2∡C+20=180 ⇒ ∡C=((180−20)/2) =80 (4)⇒((sin 20)/(sin (θ−20))) =((sin 80)/(sin θ)) sin 20sin θ=sin 80sin (θ−20) E=sin 20sin θ−sin θcos 20sin 80+cos θsin 20sin 80=0 E×(1/(cos 20)) sin θ(tan 20−sin 80)+cos θtan 20sin 80=0 ⇒tan 𝛉=((tan 20sin 80)/(sin 80−tan 20)) tan𝛉=0,57735026.. 𝛉=30

$$\left.\bullet\mathrm{1}\right)\bigtriangleup\mathrm{ABM}\:\:\:\:\frac{\mathrm{BM}}{\mathrm{sin}\:\mathrm{20}}=\frac{\mathrm{AM}}{\mathrm{sin}\:\mathrm{B1}}=\frac{\mathrm{AB}}{\mathrm{sin}\:\left(\pi−\theta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{AMsin}\:\mathrm{20}=\mathrm{BMsin}\:\mathrm{B1}\:\:\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{ABsin}\:\mathrm{20}=\mathrm{BMsin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{BM}=\frac{\mathrm{ABsin}\:\mathrm{20}}{\mathrm{sin}\:\theta}\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left.\bullet\mathrm{2}\right)\bigtriangleup\mathrm{BMC}\:\:\:\:\frac{\mathrm{BM}}{\mathrm{sin}\:\mathrm{C}}=\frac{\mathrm{BC}}{\mathrm{sin}\:\theta}=\frac{\mathrm{MC}}{\mathrm{sin}\:\left(\mathrm{B}−\mathrm{B1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{BMsin}\:\theta=\mathrm{BCsin}\:\mathrm{C}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\theta=\mathrm{20}+\mathrm{B1}\Rightarrow\mathrm{B1}=\theta−\mathrm{20} \\ $$$$\:\:\:\:\:\:\left(\mathrm{1}\right)\Rightarrow\mathrm{BM}=\frac{\mathrm{AMsin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{B1}}=\frac{\mathrm{AMsin}\:\mathrm{20}}{\mathrm{sin}\:\left(\theta−\mathrm{20}\right)} \\ $$$$\:\:\:\:\:\:\left(\mathrm{3}\right)\Rightarrow\mathrm{BM}=\frac{\mathrm{BCsin}\:\mathrm{C}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{AMsin}\:\mathrm{20}}{\mathrm{sin}\:\left(\theta−\mathrm{20}\right)}=\frac{\mathrm{BCsin}\:\mathrm{C}}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\mathrm{AM}=\mathrm{BC}\Rightarrow\:\:\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\left(\boldsymbol{\theta}−\mathrm{20}\right)}=\frac{\mathrm{sin}\:\boldsymbol{\mathrm{C}}}{\mathrm{sin}\:\boldsymbol{\theta}}\:\:\left(\mathrm{4}\right) \\ $$$$\left.\bullet\mathrm{3}\right)\bigtriangleup\mathrm{ABC}\:\:\:\:\mathrm{AC}=\mathrm{AB} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\:\measuredangle\mathrm{B}=\measuredangle\mathrm{C}\:\:;\:\:\:\mathrm{2}\measuredangle\mathrm{C}+\mathrm{20}=\mathrm{180} \\ $$$$\:\:\:\:\:\Rightarrow\:\:\measuredangle\mathrm{C}=\frac{\mathrm{180}−\mathrm{20}}{\mathrm{2}}\:\:\:=\mathrm{80} \\ $$$$\left(\mathrm{4}\right)\Rightarrow\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\left(\theta−\mathrm{20}\right)}\:\:=\frac{\mathrm{sin}\:\mathrm{80}}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\mathrm{20sin}\:\theta=\mathrm{sin}\:\mathrm{80sin}\:\left(\theta−\mathrm{20}\right) \\ $$$$\:\:\:\mathrm{E}=\mathrm{sin}\:\mathrm{20sin}\:\theta−\mathrm{sin}\:\theta\mathrm{cos}\:\mathrm{20sin}\:\mathrm{80}+\mathrm{cos}\:\theta\mathrm{sin}\:\mathrm{20sin}\:\mathrm{80}=\mathrm{0} \\ $$$$\:\:\:\mathrm{E}×\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}} \\ $$$$\:\:\:\mathrm{sin}\:\theta\left(\mathrm{tan}\:\mathrm{20}−\mathrm{sin}\:\mathrm{80}\right)+\mathrm{cos}\:\theta\mathrm{tan}\:\mathrm{20sin}\:\mathrm{80}=\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\Rightarrow\mathrm{tan}\:\boldsymbol{\theta}=\frac{\boldsymbol{\mathrm{tan}}\:\mathrm{20}\boldsymbol{\mathrm{sin}}\:\mathrm{80}}{\boldsymbol{\mathrm{sin}}\:\mathrm{80}−\boldsymbol{\mathrm{tan}}\:\mathrm{20}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{tan}\theta}=\mathrm{0},\mathrm{57735026}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\theta}=\mathrm{30} \\ $$$$\:\:\:\:\:\:\: \\ $$

Commented by Mingma last updated on 18/Apr/23

Excellent!

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