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In parallelogram ABCD,AB^4+AD^2+AB^2*AD^2=AC^2*BD^2.If angle ABC=x,find the product of all possible values of x

Sagot :

Caylusidn

Answer:

Hello,

Step-by-step explanation:

AB=DC=b, AD=BC=a

p=BD, q=AC

angle ABC=x

[tex]AB^4+AD^4+AB^2*AD^2=AC^2*BD^2\\\\b^4+a^4+b^2a^2=(a^2+b^2+2abcos(X))(a^2+b^2-2abcos(X)\\\\(a^2+b^2+2a^2b^2)-a^2b^2=((a^2+b^2)^2-4a^2b^2*cos^2(X))\\\\\\4a^2b^2cos^2(X)=a^2b^2\\\\\\cos^2(X)=\dfrac{1}{4} \\\\(cos(X)-\dfrac{1}{2} )(cos(X)+\dfrac{1}{2} )=0\\cos(x)=\frac{1}{2} \ or\ cos(X)=\dfrac{-1}{2} \\\\\\X=60^o =\dfrac{\pi}{3} rad \ or \ X=120^o=\dfrac{4\pi}{3} rad\\\\\\\\Product=\dfrac{\pi}{3}*\dfrac{4\pi}{3} =\boxed{\dfrac{4\pi^2}{9}}\\[/tex]

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