(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Imagine an isoscoles triangle. The "top" vertex has an angle of 2*theta. The common side has a length of L. When you cut the isoscoles triangle in half, that common side becomes the hypotenuse of the two resulting right-triangles.

[itex]{\rm{BaseLength}} = L \cdot \sin \theta + L\sin \theta = 2L\sin \theta [/itex]

[itex]{\rm{HeightLength}} = L\cos \theta [/itex]

I also am under the impression that:

[itex]{\rm{Area = BaseLength}} \times {\rm{HeightLength}}[/itex]

...so obviously:

[itex]A = 2{L^2}\sin \theta \cos \theta [/itex]

If this is the correct area, then I've found a typo in a book... Can you either confirm my suspicions or refute my stance?

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# Homework Help: Area of an isoscoles triangle

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