Question

A parallelogram has sides of equal length but the distance between two opposite vertices 1.5 times that of the length of a given side. Use the Sine rule to calculate the interior angle of the parallelogram?

confused about this question

Answer 1

Given that ABCD is a parallelogram which has sides of equal length. So ABCD is a rhombus. Hence its diagonals will bisect each other perpendicularly.

So `AB=BC=CD=DA` Also iis given that `(BD)/(AB)=1.5`

Now for right `DeltaAOB`

`cosangleABO =(OB)/(AB)=1/2xx(BD)/(AB)=1.5/2=0.75`

`=>angle ABO=angle ABD=cos^-1(0.75)=41.4^@`

Hence `angle ADC=angle ABC=2angle ABD=2xx41,4^@=82.8^@`

Obviously `angle BCD=angle BAD=180^@-82.8^@=97.2^@`