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Question Number 149932 by mathdanisur last updated on 08/Aug/21

Answered by mr W last updated on 08/Aug/21

say AB=CD=1 ((BD)/(sin 5x))=((BC)/(sin 9x))=(1/(sin 4x)) ⇒BD=((sin 5x)/(sin 4x)) ⇒BC=((sin 9x)/(sin 4x)) ((AD)/(sin 25x))=(1/(sin 9x))=((BD)/(sin 16x)) ⇒AD=((sin 25)/(sin 9x)) ⇒BD=((sin 16x)/(sin 9x)) ((sin 16x)/(sin 9x))=((sin 5x)/(sin 4x)) ...(i) ((BC)/(sin 16x))=(1/(sin 5x)) ⇒BC=((sin 16x)/(sin 5x)) ((sin 16x)/(sin 5x))=((sin 9x)/(sin 4x)) ...(ii)≡(i) ⇒x≈6°

$${say}\:{AB}={CD}=\mathrm{1} \\ $$$$\frac{{BD}}{\mathrm{sin}\:\mathrm{5}{x}}=\frac{{BC}}{\mathrm{sin}\:\mathrm{9}{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{9}{x}}{\mathrm{sin}\:\mathrm{4}{x}} \\ $$$$\frac{{AD}}{\mathrm{sin}\:\mathrm{25}{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{9}{x}}=\frac{{BD}}{\mathrm{sin}\:\mathrm{16}{x}} \\ $$$$\Rightarrow{AD}=\frac{\mathrm{sin}\:\mathrm{25}}{\mathrm{sin}\:\mathrm{9}{x}} \\ $$$$\Rightarrow{BD}=\frac{\mathrm{sin}\:\mathrm{16}{x}}{\mathrm{sin}\:\mathrm{9}{x}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{16}{x}}{\mathrm{sin}\:\mathrm{9}{x}}=\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{4}{x}}\:\:\:...\left({i}\right) \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{16}{x}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{5}{x}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{sin}\:\mathrm{16}{x}}{\mathrm{sin}\:\mathrm{5}{x}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{16}{x}}{\mathrm{sin}\:\mathrm{5}{x}}=\frac{\mathrm{sin}\:\mathrm{9}{x}}{\mathrm{sin}\:\mathrm{4}{x}}\:\:\:...\left({ii}\right)\equiv\left({i}\right) \\ $$$$\Rightarrow{x}\approx\mathrm{6}° \\ $$

Commented by mathdanisur last updated on 08/Aug/21

Ser, Thank You

$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Aug/21

△ABC: ∠ABD=π−(16x+5x+4x) =π−25x △ABD: ∠ADB=π−16x−(π−25x) =9x △BDC: ∠BDC=π−(4x+5x) =π−9x ∠ADB+∠CDB=π 9x+π−9x=π π=π free of x Any value of x for which each of all angles fall between 0 & π.

$$\bigtriangleup{ABC}:\:\angle{ABD}=\pi−\left(\mathrm{16}{x}+\mathrm{5}{x}+\mathrm{4}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pi−\mathrm{25}{x} \\ $$$$\bigtriangleup{ABD}:\:\angle{ADB}=\pi−\mathrm{16}{x}−\left(\pi−\mathrm{25}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}{x} \\ $$$$\bigtriangleup{BDC}:\:\angle{BDC}=\pi−\left(\mathrm{4}{x}+\mathrm{5}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pi−\mathrm{9}{x} \\ $$$$\angle{ADB}+\angle{CDB}=\pi \\ $$$$\mathrm{9}{x}+\pi−\mathrm{9}{x}=\pi \\ $$$$\:\:\:\:\:\:\:\:\pi=\pi\:\:\:{free}\:{of}\:{x} \\ $$$${Any}\:{value}\:{of}\:{x}\:{for}\:{which}\:{each}\:{of} \\ $$$${all}\:{angles}\:{fall}\:{between}\:\mathrm{0}\:\&\:\pi. \\ $$

Commented by mathdanisur last updated on 08/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

Commented by prakash jain last updated on 08/Aug/21

AB=CD will not satisfy for all x.

$${AB}={CD}\:\mathrm{will}\:\mathrm{not}\:\mathrm{satisfy}\:\mathrm{for}\:\mathrm{all}\:{x}. \\ $$

Commented by Rasheed.Sindhi last updated on 09/Aug/21

ThanX Sir prakash jain, Actually I didn′t see it and therefore didn′t consider it.

$$\mathcal{T}{han}\mathcal{X}\:\:{Sir}\:{prakash}\:{jain},\:{Actually}\:{I}\:{didn}'{t} \\ $$$${see}\:{it}\:{and}\:{therefore}\:{didn}'{t}\:{consider} \\ $$$${it}. \\ $$

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