Question

$\underline\mathtt{Question}$

Two sides of a triangle are given find the angle between them such that the area of the triangle is maximum.

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Answer 1

Answer:

Let a and b be the lengths of the given sides and let θ be the angle between them.

Let A be the area of the triangle.

$a = \frac{1}{2} ab \sin( θ)$

$a(θ) = \frac{1}{2} ab \ \sin( \:θ)$

${ a( θ) = \frac{1}{2} ab \cos( \:θ)$

$a(θ) = \frac{ - 1}{2} ab \sin( θ)$

$a(θ) = 0$

$θ = \frac{\pi}{2}$

$a \frac{\pi}{2} = \frac{ - 1}{2} ab \sin( \frac{\pi}{2} ) = \frac{ - 1}{2} ab < 0$

$θ = \frac{\pi}{2}$

Answer 2

Answer :

π/2

Solution :

Consider a ∆ABC with BC = a and AC = b be the two given sides Let θ be the angle between BC and AC . Let's draw the perpendicular BD = p on the side AC from the vertex B .

Now ,

In ∆BCD ,

=> sinθ = BD/BC

=> sinθ = p/a

=> p = a•sinθ

Also ,

We know that , the area of a triangle is given as : A = ½•Base•Height

Thus ,

The area of ∆ABC will be ;

=> A = ½•Base•Height

=> A = ½•b•a•sinθ

=> A(θ) = ½•ab•sinθ

Clearly ,

The area A is a function of angle θ .

Now ,

Differentiating both the sides with respect to θ , we get ;

=> dA/dθ = d(½•ab•sinθ)/dθ

=> dA/dθ = ½•ab•d(sinθ)/dθ

=> A'(θ) = ½•ab•cosθ

Also ,

We know that , if f'(x) = 0 at x = a , then f(x) is either maximum or minimum at x = a .

Thus ,

For the area A(θ) to be maximum ,

A'(θ) = 0 .

=> ½•ab•cosθ = 0

=> cosθ = 0

=> cosθ = cos(π/2)

=> θ = π/2

Hence ,

The area of a triangle will be maximum when the angle between the two given sides is π/2 .