Question #143551
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
1
Expert's answer
2020-11-17T16:57:20-0500

Let's say one side's length is a. Enclosed area is a triangle with a2\frac{a}{2} as a side opposed to α\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 π3\frac{\pi}{3} radian (as the triangle is equal-sided), so a third angle would be ππ3α=2π3α\pi-\frac{\pi}{3}-\alpha=\frac{2\pi}{3}-\alpha radian.


Let's say that a side opposite to π3\frac{\pi}{3} angle is of length x and opposite to 2π3α\frac{2\pi}{3}-\alpha is of length y.


a2sinα=xsin(π3)=ysin(2π3α)\frac{\frac{a}{2}}{sin\alpha}=\frac{x}{sin(\frac{\pi}{3})}=\frac{y}{sin(\frac{2\pi}{3}-\alpha)}


x=a2sin(π3)sinα,y=a2sin(2π3α)sinαx=\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 12absinA\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=12xysinα=a8sinαsin(π3)sin(2π3α)sin2α=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}=

=a8sin(π3)sin(2π3α)sinα=\frac{a}{8}\cdot\frac{sin(\frac{\pi}{3})\cdot sin(\frac{2\pi}{3}-\alpha)}{sin\alpha} .


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