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
Answers
Given : 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.
To Find : Derive an expression for the enclosed area A(α) with respect to the angle
Solution:
ABC is a triangle , AB = BC = AC = a
D is center of BC => BD = CD = a/2
case 1 - α is between 0 and 90 deg
DE meets AC at E making angle α
A(α) = Area of Triangle DCE
∠DCE = 60°
∠CDE = α
∠DEC = 120° - α
Using Sine rule of triangle
CD/Sin(120° - α ) = CE/Sinα = DE/Sin60°
=> a/2Sin(120° - α ) = CE/Sinα = DE/Sin60°
=> DE = aSin60°/2Sin(120° - α )
A(α) = Area of Triangle DCE = (1/2) CD. DE.Sin α
= (1/2) (a/2). DE.Sin α
= (1/2) (a/2). (aSin60°/2Sin(120° - α )).Sin α
Sin60° = √3/2
= (√3a²/16)Sin α /Sin(120° - α )
Sin(120° - α ) = Sin(60° + α )
= (√3 a²/16)Sin α /Sin(60° + α )
A(α) = (√3 a²/16)Sin α /Sin(60° + α ) for α is between 0 and 90°
case 2 - α is between 90 and 180°
DF meets AB at F making angle α
A(α) = Area of Triangle ABC - Area of Triangle BDF
∠DBF = 60°
∠BDF = 180°- α
∠DFB = α - 60°
Using Sine rule of triangle
BD/Sin(α - 60°) = BF/Sin(180°- α) = DF/Sin60°
=> a/2Sin(α - 60°) = DF/Sin60°
=> DE = aSin60°/2Sin(α - 60°)
A(α) = Area of Triangle ABC - Area of Triangle BDF
= (√3 /4 )a² - (1/2)BD. DF Sin(180°- α)
= (√3 /4 )a² - (1/2) (a/2). DF.Sin α
= (√3 /4 )a² - (1/2) (a/2). (aSin60°/2Sin(α - 60°)).Sin α
Sin60° = √3/2
= (√3 /4 )a² - (√3a²/16)Sin α /Sin(α - 60°)
for α between 90 and 180 deg
A(α) = (√3 a²/16)Sin α /Sin(60° + α ) for α is between 0 and 90°
(√3 /4 )a² - (√3a²/16)Sin α /Sin(α - 60°) for α between 90° and 180°
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Derive an expression for the enclosed area A(α) with respect to the angle
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