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.
[Math] How to find the area of an isosceles triangle without using trigonometry
areaeuclidean-geometrygeometrytriangles
Related Solutions
Use:
In triangle $ABD:$
$$\angle D=90^{\circ},\angle A=30^{\circ} \Rightarrow BD=\frac12AB=10$$
Let $QM$ be an altitude of $\Delta PQB$.
Thus, $$MQ=QC=\frac{1}{2}PQ,$$ $$\measuredangle MPQ=30^{\circ},$$ $$\measuredangle PAQ=15^{\circ}.$$ Can you end it now?
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Best Answer
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:
$$S=\frac{10 \cdot 10 \cdot c}{4R}=\frac{100}{4}=25$$