The area of a triangle by the sine

The area of a triangle is half a side times the height to it. When two sides and the angle between them are known, the sine gives the height:

The proof below draws the height $\displaystyle {BD}={h}$ to the side $\displaystyle b$, and shows that $\displaystyle {h}={a\sin\gamma}$.

• In the right-angled triangle $\displaystyle BDC$ the hypotenuse is $\displaystyle a$, and $\displaystyle h$ is the leg opposite $\displaystyle \gamma$: $\displaystyle {\sin\gamma}={\frac{h}{a}}$, that is $\displaystyle {h}={a\sin\gamma}$.
• So $\displaystyle {S}={\frac{1}{2}b\cdot h}={\frac{1}{2}ab\sin\gamma}$.

When $\displaystyle \gamma$ is obtuse, the height from $\displaystyle B$ meets the extension of the side $\displaystyle b$, outside the triangle.

• In the right-angled triangle $\displaystyle BDC$ the angle at $\displaystyle C$ is $\displaystyle 180^{\circ}-\gamma$ (an adjacent angle), so $\displaystyle {h}={a\sin(180^{\circ}-\gamma)}$.
• $\displaystyle {\sin(180^{\circ}-\gamma)}={\sin\gamma}$, so again $\displaystyle {h}={a\sin\gamma}$ and $\displaystyle {S}={\frac{1}{2}ab\sin\gamma}$.

When $\displaystyle {\gamma}={90^{\circ}}$, $\displaystyle {\sin\gamma}={1}$ and the formula is half the product of the legs. The formula holds in every triangle.

• The best angle: when the sides $\displaystyle a$ and $\displaystyle b$ are fixed, the area is largest when $\displaystyle {\sin\gamma}={1}$, that is $\displaystyle {\gamma}={90^{\circ}}$.
• Two angles, one area: $\displaystyle {\sin(180^{\circ}-\gamma)}={\sin\gamma}$, so a triangle with the angle $\displaystyle 40^{\circ}$ between the sides $\displaystyle a$ and $\displaystyle b$, and a triangle with the angle $\displaystyle 140^{\circ}$ between the same sides, have the same area.
• A parallelogram with sides $\displaystyle a$, $\displaystyle b$ and the angle $\displaystyle \alpha$ between them: a diagonal cuts it into two congruent triangles, so its area is $\displaystyle ab\sin\alpha$.

Work out the area, as an exact value.

A triangle: $\displaystyle {a}={8}$, $\displaystyle {b}={6}$, and the angle between them is $\displaystyle 30^{\circ}$.

A triangle: $\displaystyle {a}={5}$, $\displaystyle {b}={8}$, and the angle between them is $\displaystyle 120^{\circ}$.

A parallelogram with sides $\displaystyle 7$ and $\displaystyle 4$, and the angle between them $\displaystyle 45^{\circ}$.

Two sides of a triangle are $\displaystyle 6$ and $\displaystyle 10$, and its area is $\displaystyle 15$.
$\displaystyle x$ is the angle between the two sides. Find all the possibilities.

Can one angle of a triangle grow by $\displaystyle 60^{\circ}$, the two sides beside it stay the same length, and the area of the triangle not change?

$\displaystyle x$ is the angle before the change. Solve the equation the equal areas give.

What are the angles before and after the change? Explain with $\displaystyle {\sin(180^{\circ}-\alpha)}={\sin\alpha}$.