Question-169947

Question Number 169947 by cortano1 last updated on 13/May/22

Commented by cortano1 last updated on 13/May/22

Answered by Rasheed.Sindhi last updated on 13/May/22

BD=1(say)⇒AC=2 sin20=((BC)/(AC)) =((BC)/2) BC=2sin20 ∠ACB=90−20=70 ∠BCD=180−70=110 △BCD: BD=1,BC=2sin20 ((BD)/(sin110))=((CD)/(sinx ))=((BC)/(sin(180−110−x) )) (1/(sin110 ))=((CD)/(sinx ))=((2sin20)/(sin(70−x) )) CD=((sinx)/(sin110 )) CD=((2sin20sinx)/(sin(70−x) )) ((sinx)/(sin110 ))=((2sin20sinx)/(sin(70−x) )) (1/(sin110 ))=((2sin20)/(sin(70−x) )) sin(70−x)=2sin20sin110 70−x=sin^(−1) (2sin20sin110) x=70−sin^(−1) (2sin20sin110) x=30

$${BD}=\mathrm{1}\left({say}\right)\Rightarrow{AC}=\mathrm{2} \\ $$$$\mathrm{sin20}=\frac{{BC}}{{AC}}\:=\frac{{BC}}{\mathrm{2}} \\ $$$${BC}=\mathrm{2sin20}\: \\ $$$$\angle{ACB}=\mathrm{90}−\mathrm{20}=\mathrm{70} \\ $$$$\angle{BCD}=\mathrm{180}−\mathrm{70}=\mathrm{110} \\ $$$$\bigtriangleup{BCD}: \\ $$$${BD}=\mathrm{1},{BC}=\mathrm{2sin20} \\ $$$$\frac{{BD}}{\mathrm{sin110}}=\frac{{CD}}{\mathrm{sin}{x}\:}=\frac{{BC}}{\mathrm{sin}\left(\mathrm{180}−\mathrm{110}−{x}\right)\:} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin110}\:}=\frac{{CD}}{\mathrm{sin}{x}\:}=\frac{\mathrm{2sin20}}{\mathrm{sin}\left(\mathrm{70}−{x}\right)\:} \\ $$$${CD}=\frac{\mathrm{sin}{x}}{\mathrm{sin110}\:} \\ $$$${CD}=\frac{\mathrm{2sin20sin}{x}}{\mathrm{sin}\left(\mathrm{70}−{x}\right)\:} \\ $$$$\frac{\mathrm{sin}{x}}{\mathrm{sin110}\:}=\frac{\mathrm{2sin20sin}{x}}{\mathrm{sin}\left(\mathrm{70}−{x}\right)\:} \\ $$$$\frac{\mathrm{1}}{\mathrm{sin110}\:}=\frac{\mathrm{2sin20}}{\mathrm{sin}\left(\mathrm{70}−{x}\right)\:} \\ $$$$\mathrm{sin}\left(\mathrm{70}−{x}\right)=\mathrm{2sin20sin110} \\ $$$$\mathrm{70}−{x}=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2sin20sin110}\right) \\ $$$${x}=\mathrm{70}−\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2sin20sin110}\right) \\ $$$${x}=\mathrm{30} \\ $$

Commented by cortano1 last updated on 13/May/22

(1/(sin 110°)) = ((2sin 20°)/(sin (70°−x))) ⇒sin (70°−x)=2sin 20° cos 20° ⇒sin (70°−x)=sin 40° ⇒x=30°

$$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{110}°}\:=\:\frac{\mathrm{2sin}\:\mathrm{20}°}{\mathrm{sin}\:\left(\mathrm{70}°−{x}\right)} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{70}°−{x}\right)=\mathrm{2sin}\:\mathrm{20}°\:\mathrm{cos}\:\mathrm{20}° \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{70}°−{x}\right)=\mathrm{sin}\:\mathrm{40}° \\ $$$$\Rightarrow{x}=\mathrm{30}° \\ $$

Answered by mr W last updated on 13/May/22

BD=1, AC=2 tan 20=((cos x)/(2 cos 20+sin x)) ((sin 20)/(cos 20))=((cos x)/(2 cos 20+sin x)) cos x cos 20−sin x sin 20=2 cos 20 sin 20 cos (x+20)=sin 40=cos 50 ⇒x+20=50 ⇒x=30 ✓

$${BD}=\mathrm{1},\:{AC}=\mathrm{2} \\ $$$$\mathrm{tan}\:\mathrm{20}=\frac{\mathrm{cos}\:{x}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}+\mathrm{sin}\:{x}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{cos}\:\mathrm{20}}=\frac{\mathrm{cos}\:{x}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{20}+\mathrm{sin}\:{x}} \\ $$$$\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{20}−\mathrm{sin}\:{x}\:\mathrm{sin}\:\mathrm{20}=\mathrm{2}\:\mathrm{cos}\:\mathrm{20}\:\mathrm{sin}\:\mathrm{20} \\ $$$$\mathrm{cos}\:\left({x}+\mathrm{20}\right)=\mathrm{sin}\:\mathrm{40}=\mathrm{cos}\:\mathrm{50} \\ $$$$\Rightarrow{x}+\mathrm{20}=\mathrm{50} \\ $$$$\Rightarrow{x}=\mathrm{30}\:\checkmark \\ $$

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