An isosceles triangle is a triangle that has two sides of equal length L (the third side may or may not have the same length). If θ is the angle between the two sides of equal length, then the formula for the area of an isosceles triangle is: A = (L^2) * sin(θ) Find the angle θ that maximizes the area of such an isosceles triangle
The Correct Answer and Explanation is:
To maximize the area of an isosceles triangle with two equal sides of length LLL, given by the formula:A=L2⋅sin(θ)A = L^2 \cdot \sin(\theta)A=L2⋅sin(θ)
we need to find the value of θ\thetaθ that maximizes this expression.
Since L2L^2L2 is a constant (because the side length does not change), the area AAA is directly proportional to sin(θ)\sin(\theta)sin(θ). That means to maximize the area, we need to maximize sin(θ)\sin(\theta)sin(θ).
The sine function sin(θ)\sin(\theta)sin(θ) reaches its maximum value of 1 when θ=90∘\theta = 90^\circθ=90∘ or θ=π2\theta = \frac{\pi}{2}θ=2π radians.
Therefore, the angle θ\thetaθ that maximizes the area of the isosceles triangle is:θ=90∘ or π2 radians\boxed{\theta = 90^\circ \text{ or } \frac{\pi}{2} \text{ radians}}θ=90∘ or 2π radians
Explanation
The area of an isosceles triangle can be determined using the formula A=L2⋅sin(θ)A = L^2 \cdot \sin(\theta)A=L2⋅sin(θ), where LLL is the length of the two equal sides and θ\thetaθ is the angle between those sides. The expression shows that the area is proportional to the square of the side length and the sine of the included angle.
Since the length LLL is fixed, we cannot change it to increase the area. The only variable that affects the area is the angle θ\thetaθ. Therefore, to maximize the area, we must find the value of θ\thetaθ for which sin(θ)\sin(\theta)sin(θ) is the largest.
The sine function reaches its maximum value of 1 when the angle is 90∘90^\circ90∘ or π2\frac{\pi}{2}2π radians. This is a fundamental property of the sine function and occurs because sine represents the height of a point on the unit circle relative to the horizontal axis. At 90∘90^\circ90∘, the point is at the highest possible vertical position on the unit circle, hence the sine is at its peak.
When θ=90∘\theta = 90^\circθ=90∘, the two equal-length sides form a right angle, creating a right triangle. In this configuration, the triangle encloses the greatest possible area for the given side lengths. Any increase or decrease in the angle from 90∘90^\circ90∘ reduces the vertical height between the two sides, which reduces the area.
Therefore, the area is maximized when the angle between the two equal sides is exactly 90∘90^\circ90∘.
