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Math Help - isosceles triangle

  1. #1
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    isosceles triangle

    Hi everyone. I've got a problem I can't solve and would love some help.

    In an isosceles triangle, give an exact answer to the angle x, so that the area of the triangle is maximized. The angle x is adjacent to the base of the triangle.
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    sorry, editing
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  3. #3
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    Area of isosceles triangle

    Hello Burger king
    Quote Originally Posted by Burger king View Post
    Hi everyone. I've got a problem I can't solve and would love some help.

    In an isosceles triangle, give an exact answer to the angle x, so that the area of the triangle is maximized. The angle x is adjacent to the base of the triangle.
    You haven't said so, but I assume that the lengths of the equal sides remain constant as x varies. If this is so, then let's assume that they are of length a. Then:

    Height of triangle = a \sin x

    Base of triangle = 2a \cos x

    So area of triangle, \triangle = \tfrac12\text{base} \times \text{height} = a^2\sin x\cos x

    Now you can use the identity \sin2x = 2\sin x\cos x, and say

    \triangle = \tfrac12a^2\sin2x

    And \sin2x has a maximum value of 1 when x = 45^o.

    So there's your answer, x = 45^o.

    Grandad
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    Thank you kindly
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