What is the angle between two vectors given their magnitudes and the area of the parallelogram they form?

Question

A vector $A$ with magnitude $15.0$ units and a vector $B$ with magnitude $7.0$ units are oriented to form an obtuse angle. The area of the parallelogram formed by them is $20.0$ units squared. What is the angle between the two vectors?

Answer 1

Let $\alpha$ be a measure of the needed angle.

Thus, $$20=15\cdot7\sin\alpha$$ or $$\sin\alpha=\frac{4}{21}$$ and since the needed angle is obtuse, we get the answer: $$180^{\circ}-\arcsin\frac{4}{21}\approx 169.019^{\circ}$$

Answer 2

By definition of cross product

$$ | A X B | = A\, B \sin \theta $$

$$ \theta = \sin^{-1} \frac {20 }{15.\,7.} $$

Out of two possibilities

$$ \theta \approx 169^0,\, 11^0 $$

choose the first obtuse angle.