How to find the area of an isosceles triangle without using trigonometry? I have an isosceles  triangle with equal sides $10$ unit, angle between them is $30^\circ$. I need to be confirmed that the area of this triangle can be found in any method other than using any kind of trigonometry. I tried Heron's formulae, but did not get any fruitful result.
 A: Try drawing an altitude from one of the $75$ degree angles to the opposite 10-unit side.  You should be able to determine the length of this altitude using a 30-60-90 triangle.  
A: Consider circumscribed circle and it's radius $R$. By inscribed angle theorem you get, that $|c|=|R|$, where $c$ is third side of your triangle $a=b=10$. Now you have formula $\displaystyle S=\frac{abc}{4R}$, where $S$ is area of triange. So:
$$S=\frac{10 \cdot 10 \cdot c}{4R}=\frac{100}{4}=25$$ 
A: I'll follow the hint of paw88789. Consider the following triangle

We have $\vert \overline{CA} \vert = \vert \overline{CB}\vert = 10$ and $\sphericalangle{ACB} = 30^{\circ}$. Now draw the height $\overline{BH}$ through $B$. We have $\sphericalangle BHC = 90^{\circ}$, therefore $\sphericalangle CBH = 60^{\circ}$. The Triangle HBC is half an equilateral triangle and therefore $\vert\overline{BH}\vert = \frac{10}{2}=5$. 
The area of the triangle is then
$$\frac12\vert \overline{AC}\vert \cdot \vert \overline{HB}\vert = \frac12 \cdot 10 \cdot 5 = 25. $$
