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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.

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  • $\begingroup$ Welcome Math.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $LaTeX$ syntax. If you receive useful answers, consider accepting one. $\endgroup$ – Shailesh Sep 30 '15 at 8:51
  • $\begingroup$ Note that 30$^{\circ}$ is not the angle between the shorter and longer side; it is the angle between the diagonal and the longer side. Therefore, your application of $ab \sin \theta$ to assume a parallelogram area of $30 cm^2$ ($10 cm \cdot 6 cm \cdot \sin 30^{\circ}$) is inaccurate. $\endgroup$ – Tebbe Sep 30 '15 at 15:50
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6_x_10_parallelogram_30_deg_diag_and_long

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.

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