Question-197095

Question Number 197095 by Mingma last updated on 07/Sep/23

Answered by mahdipoor last updated on 07/Sep/23

((AB)/(sin4a))=((BM)/(sina)) ((CD)/(sin4a))=((MC)/(sin2a))⇒((2.CD.cosa)/(sin4a))=((MC)/(sina))⇒ ((MC)/(sina))+((BM)/(sina))=((CB)/(sina))=((2.CD.cosa)/(sin4a))+((AB)/(sin4a)) ⇒((sin4a)/(sina))=4cos2a.cosa=2cosa+1⇒ 8cos^3 a−6cosa−1=0⇒a=20 (how??)

$$\frac{{AB}}{{sin}\mathrm{4}{a}}=\frac{{BM}}{{sina}}\:\:\: \\ $$$$\:\frac{{CD}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sin}\mathrm{2}{a}}\Rightarrow\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sina}}\Rightarrow \\ $$$$\frac{{MC}}{{sina}}+\frac{{BM}}{{sina}}=\frac{{CB}}{{sina}}=\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}+\frac{{AB}}{{sin}\mathrm{4}{a}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{4}{a}}{{sina}}=\mathrm{4}{cos}\mathrm{2}{a}.{cosa}=\mathrm{2}{cosa}+\mathrm{1}\Rightarrow \\ $$$$\mathrm{8}{cos}^{\mathrm{3}} {a}−\mathrm{6}{cosa}−\mathrm{1}=\mathrm{0}\Rightarrow{a}=\mathrm{20}\:\:\:\left({how}??\right) \\ $$

Commented by sniper237 last updated on 07/Sep/23

⇒^(linearisation ) 2cos(3a)−1=0 ⇒^(trigo eqt%) cos(3a)=cos(60) Then 3a=60 and a=20

$$\overset{{linearisation}\:} {\Rightarrow}\mathrm{2}{cos}\left(\mathrm{3}{a}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\overset{{trigo}\:{eqt\%}} {\Rightarrow}\:{cos}\left(\mathrm{3}{a}\right)={cos}\left(\mathrm{60}\right) \\ $$$${Then}\:\:\mathrm{3}{a}=\mathrm{60}\:{and}\:{a}=\mathrm{20} \\ $$

Commented by Mingma last updated on 08/Sep/23

Perfect ��

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Question-197095

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