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Question Number 89496 by I want to learn more last updated on 17/Apr/20

	Answered by me2love2math last updated on 17/Apr/20

	$u s e s i n e r u l e$
	Answered by $@ty@m123 last updated on 17/Apr/20
	$((BD)/(sin {180−(45+x)}))=((AD)/(sin x)) ((BD)/(sin(45+x)))=((AD)/(sin x)) ((BD (√2))/(cos x+sin x))=((AD)/(sin x)) ((BD)/(AD))=((cos x+sin x)/((√2)sin x)) ...(1) ((BD)/(sin x))=((AD)/(sin {45−x))) {∵x+∠C =45^o ((BD)/(sin x))=((AD(√2))/(cos x−sin x)) ((BD)/(AD))=(((√2)sin x)/(cos x−sin x)) ....(2) From (1) & (2), ((cos x+sin x)/((√2)sin x))=(((√2)sin x)/(cos x−sin x)) cos^2 x−sin^2 x=2sin^2 x 3sin^2 x=cos^2 x tan^2 x=(1/3) x=30^o$
	$\frac{B D}{\sin {180 - (45 + x)}} = \frac{A D}{\sin x}$ $\frac{B D}{\sin (45 + x)} = \frac{A D}{\sin x}$ $\frac{B D \sqrt{2}}{\cos x + \sin x} = \frac{A D}{\sin x}$ $\frac{B D}{A D} = \frac{\cos x + \sin x}{\sqrt{2} \sin x} . . . (1)$ $\frac{B D}{\sin x} = \frac{A D}{\sin {45 - x)} {∵ x + ∠ C = 45^{o}$ $\frac{B D}{\sin x} = \frac{A D \sqrt{2}}{\cos x - \sin x}$ $\frac{B D}{A D} = \frac{\sqrt{2} \sin x}{\cos x - \sin x} . . . . (2)$ $F r o m (1) & (2),$ $\frac{\cos x + \sin x}{\sqrt{2} \sin x} = \frac{\sqrt{2} \sin x}{\cos x - \sin x}$ $\cos^{2} x - \sin^{2} x = 2 \sin^{2} x$ $3 \sin^{2} x = \cos^{2} x$ $\tan^{2} x = \frac{1}{3}$ $x = 30^{o}$
	Commented by I want to learn more last updated on 17/Apr/20

	$Thanks sir .$
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