Question

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

Answer 1

Hello,

Explanation:

AB=DC=b, AD=BC=a

p=BD, q=AC

angle ABC=x

`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)=(1)/(4) \\\\(cos(X)-(1)/(2) )(cos(X)+(1)/(2) )=0\\cos(x)=(1)/(2) \ or\ cos(X)=(-1)/(2) \\\\\\X=60^o =(\pi)/(3) rad \ or \ X=120^o=(4\pi)/(3) rad\\\\\\\\Product=(\pi)/(3)*(4\pi)/(3) =\boxed{(4\pi^2)/(9)}\\`