Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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

$$\mathrm{nice} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com