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.
In the picture shown, $AB = DC = 10$, $BC = AD = 6$, and $\angle BDC = 30^{\circ}$.
Apply the Law of Cosines to $\triangle BDC$. Then
$$6^2 = d^2 + 10^2 - 2 \cdot d \cdot 10 \cdot \cos 30^{\circ}$$ $$36 = d^2 + 100 - 20d \cdot \frac{\sqrt{3}}{2}$$ $$0 = d^2 - 10d\sqrt{3} + 64$$ $$d^2=10d\sqrt{3} + 64$$ $$d=\sqrt{10d\sqrt{3} + 64}$$
Apply the quadratic formula to solve for the long diagonal $d$. You should get two (geometrically valid) roots, although only one of the roots makes $BD$ the long diagonal; the other root makes it the short one.