Question-89496

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

Answered by me2love2math last updated on 17/Apr/20

$${use}\:{sine}\:{rule} \\ $$

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

$$\frac{{BD}}{\mathrm{sin}\:\left\{\mathrm{180}−\left(\mathrm{45}+{x}\right)\right\}}=\frac{{AD}}{\mathrm{sin}\:{x}} \\ $$$$\frac{{BD}}{\mathrm{sin}\left(\mathrm{45}+{x}\right)}=\frac{{AD}}{\mathrm{sin}\:{x}} \\ $$$$\frac{{BD}\:\sqrt{\mathrm{2}}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:{x}}\: \\ $$$$\frac{{BD}}{{AD}}=\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}\:…\left(\mathrm{1}\right) \\ $$$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}}{\mathrm{sin}\:\left\{\mathrm{45}−{x}\right)}\:\left\{\because{x}+\angle{C}\:=\mathrm{45}^{\mathrm{o}} \right. \\ $$$$\frac{{BD}}{\mathrm{sin}\:{x}}=\frac{{AD}\sqrt{\mathrm{2}}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\: \\ $$$$\frac{{BD}}{{AD}}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\:….\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right), \\ $$$$\frac{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}=\mathrm{2sin}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{3sin}\:^{\mathrm{2}} {x}=\mathrm{cos}\:^{\mathrm{2}} {x} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}=\mathrm{30}^{\mathrm{o}} \\ $$

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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