Hello, Crowdia!

3. Two sides of a triangle have lengths $\displaystyle a$ and $\displaystyle b$, and the angle between them is $\displaystyle \theta$.

What value of $\displaystyle \theta$ will maximize the triangle's area?

Hint: .$\displaystyle A \:=\:\tfrac{1}{2}ab\sin\theta$ Differentiate and equate to zero . . .

. . $\displaystyle \frac{dA}{d\theta} \:=\: \tfrac{1}{2}ab\cos\theta \:=\:0 \quad\Rightarrow\quad \cos\theta \:=\:0 \quad\Rightarrow\quad \theta \:=\:\frac{\pi}{2} \:=\:90^o$

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With a little Thought, you can "eyeball" the solution. Code:

*
/: .
a / : .
/ :h .
/ : .
/ θ : .
* - - + - - - - - *
: - - - b - - - :

Since $\displaystyle A \:=\:\tfrac{1}{2}bh$, the area is a maximum when $\displaystyle h$ is at its maximum.

And this happens when $\displaystyle h = a:\;\theta \,=\,90^o.$