You know $\text{Area} = \frac{1}{2}bh$ from Area of a Triangle. If you know two sides and the angle between them, but not the height, the trig from Solving Right Triangles supplies the height.
Why it works
In $\triangle ABC$, take $a = BC$ as the base. Drop the height $h$ from $A$ to line $BC$. That makes a right triangle with hypotenuse $b = CA$ and the angle $C$, so $\sin C = \frac{h}{b}$ and $h = b \sin C$. Then the area is $\frac{1}{2} \cdot a \cdot b \sin C$. If $C$ is obtuse, the height falls outside the triangle and the right triangle uses the angle $180^\circ - C$ instead; since $\sin(180^\circ - C) = \sin C$ (your calculator agrees), the same formula works.
Key idea
The area of a triangle is half the product of two sides and the sine of the angle between them: $\text{Area} = \frac{1}{2}ab \sin C$.
Worked example: An exact area
Two sides of a triangle are $8$ and $10$, and the angle between them is $30^\circ$. Find the area.
- $\text{Area} = \frac{1}{2} \cdot 8 \cdot 10 \cdot \sin 30^\circ = 40 \cdot \frac{1}{2} = 20$.
Answer: $20$ square units
Worked example: With a calculator
A triangular garden has sides $12$ m and $15$ m that meet at $72^\circ$. Find its area to the nearest tenth.
- $\text{Area} = \frac{1}{2} \cdot 12 \cdot 15 \cdot \sin 72^\circ = 90 \sin 72^\circ$, which is about $90(0.9511) \approx 85.6$.
Answer: About $85.6$ m$^2$
Worked example: An obtuse angle
Find the exact area of a triangle with sides $6$ and $7$ and an included angle of $120^\circ$.
- $\sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$.
- $\text{Area} = \frac{1}{2} \cdot 6 \cdot 7 \cdot \frac{\sqrt{3}}{2} = \frac{21\sqrt{3}}{2}$.
Answer: $\frac{21\sqrt{3}}{2} \approx 18.2$
The formula also shows how the angle controls the area. With sides $a$ and $b$ fixed, the area is largest when $C = 90^\circ$, since $\sin 90^\circ = 1$, and it shrinks toward 0 as the angle closes up or opens out flat. A parallelogram with sides $a$ and $b$ and an angle $C$ is two such triangles, so its area is $ab \sin C$.
Common mistake
Using an angle that is not between the two sides. The formula needs the included angle, the one formed by the two sides you multiply. With sides $a$ and $b$, use $C$; with $b$ and $c$, use $A$.