Question-208467

Question Number 208467 by Tawa11 last updated on 16/Jun/24

Answered by A5T last updated on 17/Jun/24

Commented by A5T last updated on 17/Jun/24

$${sin}\theta=\frac{\mathrm{3}\sqrt{\mathrm{10}}}{\:\mathrm{10}}\Rightarrow{cos}\theta=\frac{\sqrt{\mathrm{10}}}{\mathrm{10}} \\ $$$${sin}\left(\mathrm{135}−\theta\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({cos}\theta+{sin}\theta\right)=\frac{\sqrt{\mathrm{20}}}{\mathrm{5}}\Rightarrow{cos}\left(\mathrm{135}−\theta\right)=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\Rightarrow{tan}\left(\mathrm{135}−\theta\right)=\mathrm{2}=\frac{\mathrm{3}}{{x}}\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\left[{blue}\right]=\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\right)×\mathrm{3}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$

Commented by Tawa11 last updated on 17/Jun/24

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

Answered by efronzo1 last updated on 17/Jun/24

Commented by Tawa11 last updated on 17/Jun/24

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

Answered by mr W last updated on 21/Jun/24

Commented by mr W last updated on 21/Jun/24

$$\sqrt{\mathrm{10}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}{x}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}{x}=\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{10}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${A}_{{blue}} =\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{Sir}.\:\:\:\:\sqrt{\mathrm{10}}\:\:=\:\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{10}}}\:\:+\:\:\frac{\mathrm{3x}}{\:\sqrt{\mathrm{10}}}\:\:\:\mathrm{not}\:\mathrm{clear}\:\mathrm{sir}. \\ $$$$\mathrm{Sorry}\:\mathrm{to}\:\mathrm{be}\:\mathrm{asking}\:\mathrm{too}\:\mathrm{much} \\ $$$$\mathrm{question}\:\mathrm{sir}. \\ $$

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