Two sides of a triangle have lengths 'a' and 'b' and the angle between them is θ. What value of θ will maximize the area of the triangle?
Solution:
The area of the triangle is given = 1/2(base)(height) --- (1)
sinθ = height/a
⇒ height = asin θ
Area of triangle(A) = (1/2)(b)(asinθ) = (1/2)absinθ
To find the angle which will maximize the area we will differentiate w.r.t. θ
dA/dθ = (1/2)ab × dsinθ/dθ
For maxima, dA/dθ = 0.
⇒ (1/2)ab cos θ = 0
Cos θ =0
⇒ θ = 90°
Therefore with θ = 90° the area of the triangle will be
A = (1/2)absin90° = (1/2)ab
Now to check that if this really the maximum we have to take a second order derivative,
d2A/dθ2 = -(1/2)absinθ which is a negative value and hence
It can be concluded that at θ = 90° the area of the triangle is maximised.
Two sides of a triangle have lengths 'a' and 'b' and the angle between them is θ. What value of θ will maximize the area of the triangle?
Summary
Given that two sides of a triangle have lengths 'a' and 'b' and the angle between them is θ. The value of θ which maximizes the area is 90° or π/2.