Answer to Question #143551 in Geometry for Vanessa

Question

You have given an equal-sided triangle with side length a. A straight line connects the center of the bottom side to the border of the triangle with an angle of α. Derive an expression for the enclosed area A(α) with respect to the angle

Answer 1

Let's say one side's length is a. Enclosed area is a triangle with $\frac{a}{2}$ as a side opposed to $\alpha$ (half of the side of the equal-sided triangle). Using law of sines, we can easily get other two sides because we have one angle $\frac{\pi}{3}$ radian (as the triangle is equal-sided), so a third angle would be $\pi-\frac{\pi}{3}-\alpha=\frac{2\pi}{3}-\alpha$ radian.

Let's say that a side opposite to $\frac{\pi}{3}$ angle is of length x and opposite to $\frac{2\pi}{3}-\alpha$ is of length y.

$\frac{\frac{a}{2}}{sin\alpha}=\frac{x}{sin(\frac{\pi}{3})}=\frac{y}{sin(\frac{2\pi}{3}-\alpha)}$

$x=\frac{a}{2}\cdot\frac{sin(\frac{\pi}{3})}{sin\alpha}, y=\frac{a}{2}\cdot\frac{sin(\frac{2\pi}{3}-\alpha)}{sin\alpha}$ .

Area is calculated as $\frac{1}{2}ab\cdot sinA$ , where a, b are sides and A is the angle between them, so in our case it is equal to:

$S=\frac{1}{2}xy\cdot sin\alpha=\frac{a}{8}\cdot\frac{sin\alpha\cdot sin(\frac{\pi}{3})\cdot sin(\frac{2\pi}{3}-\alpha)}{sin^2\alpha}=$

$=\frac{a}{8}\cdot\frac{sin(\frac{\pi}{3})\cdot sin(\frac{2\pi}{3}-\alpha)}{sin\alpha}$ .