The longer side of a parallelogram is 10 cm and shorter is 6 cm. If the longer diagonal makes an angle 30 degrees with the longer side, find the length of the longer diagonal.
In this question let $AB=10$ & $BC=6$. Height can be determined by $ab\sin\theta = base \times height$ Therefore height = 3.
If BM is height then we can find the distance MC (by Pythagoras theorem) as $3\sqrt 3$ Again using Pythagoras' theorem for AEC(ec is height) $diagonal^2 = 3^2 +(10+3\sqrt3)^2$
But this is coming out to be wrong. Please let me know if I am doing it wrong.