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Question-176950




Question Number 176950 by cortano1 last updated on 28/Sep/22
Answered by mr W last updated on 28/Sep/22
say ∠AEC=θ  CB=4 sin θ  BE=4 cos θ  ((AE)/(sin 45°))=(3/(sin (θ−45°)))  AE=(3/(sin θ−cos θ))=4 sin θ+4 cos θ  sin^2  θ−cos^2  θ=(3/4)  sin^2  θ=(7/8) ⇒sin θ=((√(14))/4)  a=4 sin θ=(√(14))  area of square =a^2 =16 sin^2  θ=14 ✓
$${say}\:\angle{AEC}=\theta \\ $$$${CB}=\mathrm{4}\:\mathrm{sin}\:\theta \\ $$$${BE}=\mathrm{4}\:\mathrm{cos}\:\theta \\ $$$$\frac{{AE}}{\mathrm{sin}\:\mathrm{45}°}=\frac{\mathrm{3}}{\mathrm{sin}\:\left(\theta−\mathrm{45}°\right)} \\ $$$${AE}=\frac{\mathrm{3}}{\mathrm{sin}\:\theta−\mathrm{cos}\:\theta}=\mathrm{4}\:\mathrm{sin}\:\theta+\mathrm{4}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta−\mathrm{cos}^{\mathrm{2}} \:\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta=\frac{\mathrm{7}}{\mathrm{8}}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{14}}}{\mathrm{4}} \\ $$$${a}=\mathrm{4}\:\mathrm{sin}\:\theta=\sqrt{\mathrm{14}} \\ $$$${area}\:{of}\:{square}\:={a}^{\mathrm{2}} =\mathrm{16}\:\mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{14}\:\checkmark \\ $$
Commented by Tawa11 last updated on 28/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by cortano1 last updated on 29/Sep/22
nice
$$\mathrm{nice} \\ $$

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