Question-204145

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Question Number 204145 by cortano12 last updated on 07/Feb/24

Answered by deleteduser1 last updated on 07/Feb/24

Commented by cortano12 last updated on 08/Feb/24

$$\:\mathrm{i}\:\mathrm{got}\:\mathrm{x}=\mathrm{40}° \\ $$

Commented by deleteduser1 last updated on 07/Feb/24

((sin(20°))/(DC))=((sin(110°))/(BC))⇒BF=DCcot(20°)−CF ((sin(110−x))/(CF))=((sin(20+x))/(DC))⇒CF=((DCsin(110−x))/(sin(20+x))) ⇒BF=DC(cot20−((sin(110−x))/(sin(20+x)))) ((sin(x))/(BF))=((sin(20))/(DF));((sin(50))/(DF))=((sin(20+x))/(DC)) ⇒DF=((DCsin(50))/(sin(20+x)))⇒((sin(x))/(BF))=((sin(20)sin(20+x))/(DCsin(50))) ⇒((sin(x)sin(20+x))/(cot20sin(20+x)−sin(110−x)))=((sin(20)sin(20+x))/(sin50)) ⇒x=20°

$$\frac{{sin}\left(\mathrm{20}°\right)}{{DC}}=\frac{{sin}\left(\mathrm{110}°\right)}{{BC}}\Rightarrow{BF}={DCcot}\left(\mathrm{20}°\right)−{CF} \\ $$$$\frac{{sin}\left(\mathrm{110}−{x}\right)}{{CF}}=\frac{{sin}\left(\mathrm{20}+{x}\right)}{{DC}}\Rightarrow{CF}=\frac{{DCsin}\left(\mathrm{110}−{x}\right)}{{sin}\left(\mathrm{20}+{x}\right)} \\ $$$$\Rightarrow{BF}={DC}\left({cot}\mathrm{20}−\frac{{sin}\left(\mathrm{110}−{x}\right)}{{sin}\left(\mathrm{20}+{x}\right)}\right) \\ $$$$\frac{{sin}\left({x}\right)}{{BF}}=\frac{{sin}\left(\mathrm{20}\right)}{{DF}};\frac{{sin}\left(\mathrm{50}\right)}{{DF}}=\frac{{sin}\left(\mathrm{20}+{x}\right)}{{DC}} \\ $$$$\Rightarrow{DF}=\frac{{DCsin}\left(\mathrm{50}\right)}{{sin}\left(\mathrm{20}+{x}\right)}\Rightarrow\frac{{sin}\left({x}\right)}{{BF}}=\frac{{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{20}+{x}\right)}{{DCsin}\left(\mathrm{50}\right)} \\ $$$$\Rightarrow\frac{{sin}\left({x}\right){sin}\left(\mathrm{20}+{x}\right)}{{cot}\mathrm{20}{sin}\left(\mathrm{20}+{x}\right)−{sin}\left(\mathrm{110}−{x}\right)}=\frac{{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{20}+{x}\right)}{{sin}\mathrm{50}} \\ $$$$\Rightarrow{x}=\mathrm{20}° \\ $$

Commented by deleteduser1 last updated on 08/Feb/24

$${Your}\:{solution}? \\ $$

Commented by A5T last updated on 27/Aug/24

((sin(20°))/(DC))=((sin(110°))/(BC))⇒BF=DCcot(20°)−CF ((sin(110−x))/(CF))=((sin(20+x))/(DC))⇒CF=((DCsin(110−x))/(sin(20+x))) ⇒BF=DC(cot20−((sin(110−x))/(sin(20+x))))...(i) ((sin(x))/(BF))=((sin(20))/(DF));((sin(50))/(DF))=((sin(20+x))/(DC)) ⇒DF=((DCsin(50))/(sin(20+x)))⇒((sin(x))/(BF))=((sin(20)sin(20+x))/(DCsin(50))) ⇒BF=((DCsin50sinx)/(sin20sin(20+x)))...(ii) (i)&(ii) ⇒sin50sinx=sin(20+x)cos20−sin(110−x)sin20 ⇒x=40°

$$\frac{{sin}\left(\mathrm{20}°\right)}{{DC}}=\frac{{sin}\left(\mathrm{110}°\right)}{{BC}}\Rightarrow{BF}={DCcot}\left(\mathrm{20}°\right)−{CF} \\ $$$$\frac{{sin}\left(\mathrm{110}−{x}\right)}{{CF}}=\frac{{sin}\left(\mathrm{20}+{x}\right)}{{DC}}\Rightarrow{CF}=\frac{{DCsin}\left(\mathrm{110}−{x}\right)}{{sin}\left(\mathrm{20}+{x}\right)} \\ $$$$\Rightarrow{BF}={DC}\left({cot}\mathrm{20}−\frac{{sin}\left(\mathrm{110}−{x}\right)}{{sin}\left(\mathrm{20}+{x}\right)}\right)…\left({i}\right) \\ $$$$\frac{{sin}\left({x}\right)}{{BF}}=\frac{{sin}\left(\mathrm{20}\right)}{{DF}};\frac{{sin}\left(\mathrm{50}\right)}{{DF}}=\frac{{sin}\left(\mathrm{20}+{x}\right)}{{DC}} \\ $$$$\Rightarrow{DF}=\frac{{DCsin}\left(\mathrm{50}\right)}{{sin}\left(\mathrm{20}+{x}\right)}\Rightarrow\frac{{sin}\left({x}\right)}{{BF}}=\frac{{sin}\left(\mathrm{20}\right){sin}\left(\mathrm{20}+{x}\right)}{{DCsin}\left(\mathrm{50}\right)} \\ $$$$\Rightarrow{BF}=\frac{{DCsin}\mathrm{50}{sinx}}{{sin}\mathrm{20}{sin}\left(\mathrm{20}+{x}\right)}…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right) \\ $$$$\Rightarrow{sin}\mathrm{50}{sinx}={sin}\left(\mathrm{20}+{x}\right){cos}\mathrm{20}−{sin}\left(\mathrm{110}−{x}\right){sin}\mathrm{20} \\ $$$$\Rightarrow{x}=\mathrm{40}° \\ $$

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