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Question-186549




Question Number 186549 by Mingma last updated on 05/Feb/23
Answered by mr W last updated on 06/Feb/23
say AD=AC=CE=a  AE=2a cos 70°  DE^2 =a^2 +(2a cos 70)^2 −2a×2a cos 70×cos 50  2^2 =a^2 (1+4 cos^2  70−4 cos 70×cos 50)  a^2 =(4/(1+4 cos^2  70−4 cos 70×cos 50))  area of ADE:  [ADE]=((a×2a cos 70×sin 50)/2)      =a^2  cos 70 sin 50      =((4 cos 70 sin 50)/(1+4 cos^2  70−4 cos 70×cos 50))      ≈1.781
$${say}\:{AD}={AC}={CE}={a} \\ $$$${AE}=\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{70}° \\ $$$${DE}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{70}\right)^{\mathrm{2}} −\mathrm{2}{a}×\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{70}×\mathrm{cos}\:\mathrm{50} \\ $$$$\mathrm{2}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{70}−\mathrm{4}\:\mathrm{cos}\:\mathrm{70}×\mathrm{cos}\:\mathrm{50}\right) \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{1}+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{70}−\mathrm{4}\:\mathrm{cos}\:\mathrm{70}×\mathrm{cos}\:\mathrm{50}} \\ $$$${area}\:{of}\:{ADE}: \\ $$$$\left[{ADE}\right]=\frac{{a}×\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{70}×\mathrm{sin}\:\mathrm{50}}{\mathrm{2}} \\ $$$$\:\:\:\:={a}^{\mathrm{2}} \:\mathrm{cos}\:\mathrm{70}\:\mathrm{sin}\:\mathrm{50} \\ $$$$\:\:\:\:=\frac{\mathrm{4}\:\mathrm{cos}\:\mathrm{70}\:\mathrm{sin}\:\mathrm{50}}{\mathrm{1}+\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{70}−\mathrm{4}\:\mathrm{cos}\:\mathrm{70}×\mathrm{cos}\:\mathrm{50}} \\ $$$$\:\:\:\:\approx\mathrm{1}.\mathrm{781} \\ $$

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