765. Two sides of a triangle are equal to $a$ and $b$, and the angle between them is $\gamma$. What is the area of this triangle
$\Delta$ The area of a triangle is half the area of the parallelogram from the previous problem (half the product of the base and height), that is, $\frac{1}{2} a b \sin \gamma \cdot \triangleleft$

Notice that in the formula for the area of a triangle, different angles (and corresponding sides) can be used. For example, for triangle $A B C$ with sides $a, b, c$ (opposite the correspondingly named angles, see the figure), we can write

$$
2 S=a b \sin \angle C=b c \sin \angle A
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-228.jpg?height=1178&width=330&top_left_y=944&top_left_x=1388)

In the last equality, $b$ can be canceled, and we get $a \sin \angle C=c \sin \angle A$. In other words, $a /(\sin \angle A)=c /(\sin \angle C)$: the ratio of a side to the sine of the opposite angle is the same for two sides. Of course, the same can be done for the other two sides, so we arrive at the equality known as the Law of Sines:

$$
\frac{a}{\sin \angle A}=\frac{b}{\sin \angle B}=\frac{c}{\sin \angle C}
$$

The Law of Sines can also be proven without using areas:

![](https://cdn.mathpix.com/cropped/2024_05_21_90703b5d5e76e3b5cd3dg-229.jpg?height=285&width=286&top_left_y=997&top_left_x=354)