Question-143909

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Question Number 143909 by mathdanisur last updated on 19/Jun/21

Commented by mr W last updated on 19/Jun/21

$${i}\:{remember}\:{i}\:{solved}\:{the}\:{same} \\ $$$${question}\:{some}\:{days}\:{ago}. \\ $$

Commented by amin96 last updated on 19/Jun/21

$$? \\ $$

Commented by mathdanisur last updated on 19/Jun/21

$${Sir},\:{if}\:{possible}\:{please}\:{take}\:{a}\:{look}\:{at}\:{this} \\ $$

Answered by mr W last updated on 19/Jun/21

Commented by mr W last updated on 19/Jun/21

say BD=DC=1 BC=2 cos 10 ((AC)/(sin 40))=((BC)/(sin 60)) AC=((2 cos 10 sin 40)/(sin 60)) ((AC)/(sin (x+10)))=((DC)/(sin x)) ((sin (x+10))/(sin x))=((2 cos 10 sin 40)/(sin 60)) cos 10+((sin 10)/(tan x))=((2 cos 10 sin 40)/(sin 60)) 1+((tan 10)/(tan x))=((4 sin 40)/( (√3))) tan x=((tan 10)/(((4 sin 40)/( (√3)))−1)) ⇒x=tan^(−1) ((tan 10)/(((4 sin 40)/( (√3)))−1))=20°

$${say}\:{BD}={DC}=\mathrm{1} \\ $$$${BC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{10} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\mathrm{40}}=\frac{{BC}}{\mathrm{sin}\:\mathrm{60}} \\ $$$${AC}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\mathrm{60}} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\left({x}+\mathrm{10}\right)}=\frac{{DC}}{\mathrm{sin}\:{x}} \\ $$$$\frac{\mathrm{sin}\:\left({x}+\mathrm{10}\right)}{\mathrm{sin}\:{x}}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\mathrm{60}} \\ $$$$\mathrm{cos}\:\mathrm{10}+\frac{\mathrm{sin}\:\mathrm{10}}{\mathrm{tan}\:{x}}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\mathrm{60}} \\ $$$$\mathrm{1}+\frac{\mathrm{tan}\:\mathrm{10}}{\mathrm{tan}\:{x}}=\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{40}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{tan}\:\mathrm{10}}{\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{40}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}} \\ $$$$\Rightarrow{x}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:\mathrm{10}}{\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{40}}{\:\sqrt{\mathrm{3}}}−\mathrm{1}}=\mathrm{20}° \\ $$

Commented by mathdanisur last updated on 19/Jun/21