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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- Jun 5th 2009, 04:34 AMBurger kingisosceles 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. - Jun 5th 2009, 04:49 AMgreat_math
sorry, editing

- Jun 5th 2009, 04:51 AMGrandadArea of isosceles triangle
Hello Burger kingYou haven't said so, but I assume that the lengths of the equal sides remain constant as $\displaystyle x$ varies. If this is so, then let's assume that they are of length $\displaystyle a$. Then:

Height of triangle = $\displaystyle a \sin x$

Base of triangle = $\displaystyle 2a \cos x$

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

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

$\displaystyle \triangle = \tfrac12a^2\sin2x$

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

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

Grandad - Jun 5th 2009, 05:07 AMBurger king
Thank you kindly :)