In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º. Find the area of ABCD.

Answer:The area of   parallelogram ABCD  is[tex]78.42 \mathrm{in}^{2}[/tex]Explanation: Given:AD = 12 in[tex]m \angle C=46^{\circ}[/tex][tex]m \angle D B A=72^{\circ}[/tex]To Find:The area of   parallelogram ABCD=?  Solution:When we construct the parallelogram with the given data, we get a parallelogram formed by 12 cm as one side and an angle with 46 degrees.  The area of the parallelogram can be calculated by [tex]a b * \sin (a n g l e)[/tex]Substituting the value of a=12  we have [tex]\text { Area of parallelogram }=12 * \text { bsin } 46[/tex]To find  the value of b, We know that area of a triangle can be expressed as, [tex]\text { Area of triangle }=(A b / 2) \sin (\text {angle})[/tex]So, [tex](12 * B D / 2) * \sin 46=(A B * B D / 2) * \sin 72[/tex]Cancelling BD and 2 on both sides we get,  [tex]12 * \sin 46=A B * \sin 72[/tex][tex]A B=12 * \frac{\sin 46}{\sin 72}[/tex]Therefore, [tex]b=\frac{12 \sin 46}{\sin 72}[/tex]Substituting the value of b,[tex]=12 *\left(\frac{12 \sin 46}{\sin 72}\right) * \sin 46[/tex]=78.42  So the area of the parallelogram is[tex]78.42 \mathrm{in}^{2}[/tex]