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Answer:
[tex]A(\alpha)=\left\{\begin{array}{ccc}\dfrac{a^2\sqrt{3}\sin{(\alpha)}}{16\sin{(\alpha+\dfrac{\pi}{3})}}&\text{if $\alpha\le\dfrac{\pi}{2}$}\\\\\dfrac{a^2\sqrt{3}}{4}\left(1-\dfrac{\sin{(\alpha)}}{4\sin{(\alpha-\dfrac{\pi}{3})}}\right)&\text{if $\alpha>\dfrac{\pi}{2}$}\end{array}\right.[/tex]
Step-by-step explanation:
If we place the bottom center at the origin and make the sides of the triangle 2 units, then the right side can be described by the equation ...
y = √3(1 -x)
and the terminal side of the angle α can be described by ...
y = tan(α)·x
We want the height of the shaded area, which is the y-coordinate of the point of intersection of these two lines.
y = √3(1 -y/tan(α)) . . . . substituting for x
y = √3·tan(α)/(√3 +tan(α)) . . . . solved for y
Since we have defined the base of the shaded area to be 1, the shaded area is half this value, or ...
A(α) = (√3/2)tan(α)/(√3 +tan(α))
Further manipulation can put this in the form ...
A(α) = (√3/4)sin(α)/sin(α+π/3) . . . . for α ≤ π/2, side length 2
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For α > π/2, we need to subtract the area A(π-α) from the area of the whole triangle. This gives an area (after some rearranging) of ...
A(α) = √3(1 -sin(α)/(4·sin(α-π/3))) . . . . for α > π/2, side length 2
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The problem statement defines the triangle as having a side length of 'a', so the final area formula will be a factor of (a/2)² times the above expressions. That is ...
[tex]A(\alpha)=\left\{\begin{array}{ccc}\dfrac{a^2\sqrt{3}\sin{(\alpha)}}{16\sin{(\alpha+\dfrac{\pi}{3})}}&\text{if $\alpha\le\dfrac{\pi}{2}$}\\\\\dfrac{a^2\sqrt{3}}{4}\left(1-\dfrac{\sin{(\alpha)}}{4\sin{(\alpha-\dfrac{\pi}{3})}}\right)&\text{if $\alpha>\dfrac{\pi}{2}$}\end{array}\right.[/tex]
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The graph (second attachment) shows the area as a function of α (in degrees) for a unit triangle. There are inflection points in the curve at 30°, 90°, and 150°
