SOLUTION: Find the angle theta (in radians) that maximizes the area of the isosceles triangle whose legs have length = 7,using the fact that the area is given by
A = 1/2l^2sin(theta).
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-> SOLUTION: Find the angle theta (in radians) that maximizes the area of the isosceles triangle whose legs have length = 7,using the fact that the area is given by
A = 1/2l^2sin(theta).
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Question 1130815: Find the angle theta (in radians) that maximizes the area of the isosceles triangle whose legs have length = 7,using the fact that the area is given by
A = 1/2l^2sin(theta). Found 2 solutions by solver91311, greenestamps:Answer by solver91311(24713) (Show Source):
You mean where is the length of one of the equal sides of the isosceles triangle and is the vertex angle.
Any single angle of any triangle must be in the range . The maximum value for the sine function in this range is 1 when . Since is independent of the measure of the vertex angle, the maximum area must occur where is maximum.
Or you can use the Calculus:
Therefore is a local extremum of the function
Which is a negative value when , therefore is a local maximum of the area function.
John
My calculator said it, I believe it, that settles it
Although not clearly stated in the problem, theta is the angle between the two sides of length 7.
If you need to use the formula to solve the problem, then consider one of the legs of length 7 as the base; then the height of the triangle is the length of the other leg of length 7, multiplied by sin(theta). So clearly the area is maximum when sin(theta) is maximum -- which is at 90 degrees.
You don't really need the formal mathematical formula to find that result. With one of the legs of length 7 the base, the height of the triangle is the distance from the end of the other leg of length 7 to the base; obviously that distance is the greatest when the angle is 90 degrees.
