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 (see drawing

Answer 1

First case;

α < 90°

The area of the shaded triangle = 1/8·a²·sin(60)sin(α)csc(120 - α)

Second case;

α > 90°

The shaded area of the triangle = 1/8·a²·sin(60)sin(180 - α)csc(60 - α) + (a²·√3)/4

Explanation:

First case;

α < 90°

The given parameters are;

The type of triangle = Equilateral triangle

The length of the sides of the triangle = a

The area of the triangle = 1/2×a×b×sinC

Where;

a and b are two legs (sides) of the triangle and C is the included angle

e/(sin(α)) = a/(2sin(120 - α)) = f/(sin60)

f = a·sin(60)/(2sin(120 - α))

The area of the triangle = 1/2 × a·sin(60)/(2sin(120 - α))×a/2 × sin(α)

By an online application, we have

1/2 × a·sin(60)/(2sin(120 - α))×a/2 × sin(α) = 1/8·a²·sin(60)sin(α)csc(120 - α)

The area of the triangle = 1/8·a²·sin(60)sin(α)csc(120 - α)

Second case;

α > 90°

The shaded area of the triangle = The area of the equilateral triangle - The unshaded area of the triangle

With an online tool, we have;

The shaded area of the triangle = (a²·√3)/4 - 1/8·a²·sin(60)sin(180 - α)csc(120 - (180 - α)

The shaded area of the triangle = 1/8·a²·sin(60)sin(180 - α)csc(60 - α) + (a²·√3)/4