Area of a Parallelogram

Question

The sides of a parallelogram measure $10$ cm and $18$ cm. One angle of the parallelogram measures $46$ degrees. What is the area of the parallelogram, to the nearest square centimeter?

I'm supposed to use the trigonometric area formula $A = \dfrac{1}{2}a b \sin C $ but I cannot seem to get it right.

Thanks in advance!

Answer 1

It should be $\dfrac{10\cdot 18}{2}\sin 46^{\circ}=90\sin 46^{\circ}\approx 65\operatorname{cm}^2$ for a triangle, so a parallelogram's area is just double this, which gives $65\cdot 2\approx 129\operatorname{cm}^2$ (after rounding)

The important thing to note is that $\dfrac{ab}{2}\sin C$ gives the area for a triangle, and a parallelogram's area is given by twice this amount: $ab\sin C.$

Answer 2

Area of a parallelogram: $A = 10 * 18 * sin (46 deg) = 180 * 0.719 = 129.48$

Area of a triangle: $A = \frac{1}{2} * 10 * 18 * sin (46 deg) = 90 * 0.719 = 64.74$