Question

You have given an equal sided triangle with side length a. A straight line connects the center

Answer 1

Answer:

Where α is an acute angle (first figure)

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

Where α is an obtuse angle (second figure)

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)

Step-by-step explanation:

Where α is an acute angle (first figure)

The given parameters are;

The given triangle = Equilateral Triangle

Let the sides of the equilateral triangle = 2·a

Therefore;

The measure of each interior angles of the given triangle = 60°

Let c represent the side of the shaded triangle opposite ∠α and b represent the side of the shaded triangle opposite ∠60° and c, represent the third side of the shaded triangle, we have;

The sides of the equilateral triangle = 2·a

By sine rule, we have;

c/sin(α) = b/sin(60°) = a/sin(180 - (60 + α)) = a/sin(120 - α))

b = sin(60°) × a/sin(120 - α)) = (√3)/2 × a/sin(120 - α))

The area of the shaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(120 - α)) × sin(α) = ((√3)·a²/4)·sin(α)·csc(120 - α))

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

Where α is an obtuse angle (second figure)

The required area of the shaded region = The area of the equilateral triangle - The area of the small unshaded triangle, with base side a and interior angles, (180° - α), 60° and ((180 - (180° - α) - 60°) = ) α - 60°

The area of the unshaded triangle is found as follows;

By sine rule, we have;

c/sin(180° - α) = b/sin(60°) = a/sin(α - 60°)

b = sin(60°) × a/sin(α - 60°) = (√3)/2 × a/sin(α - 60°)

The area of the unshaded triangle = 1/2 × a × b × sin(α) = 1/2 × a × (√3)/2 × a/sin(α - 60°) × sin(α) = -((√3)·a²/4)·sin(α)·sec(α + π/6)

The area of the shaded triangle =  -((√3)·a²/4)·sin(α)·sec(α + π/6)

The required area of the shaded region = 1/2×a²·sin(60°)  - (-((√3)·a²/4)·sin(α)·sec(α + π/6))

The required area of the shaded region = (√3)·a²/4 + (√3)·a²/4)·sin(α)·sec(α + π/6)