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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.

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    $\begingroup$ Area of a triangle without trigonometry? It is preposterous. I think you mean something else. Trigonometry is the study of relationships involving lengths and angles of triangles. $\endgroup$ – glebovg Aug 11 '14 at 1:59
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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:

$$S=\frac{10 \cdot 10 \cdot c}{4R}=\frac{100}{4}=25$$

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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.

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  • $\begingroup$ And furthermore, note that the leftover portion is similar to half the original triangle bisected at the 30 degree angle. $\endgroup$ – David H Aug 9 '14 at 13:17
  • $\begingroup$ Using your hint ,applied on $\triangle ABC$ ,where $AB=AC =10 $ cm. and $\angle BAC=30^ \circ ,BD \perp AC $, I see that $\triangle ABC= \frac 12 \times 10 \times BD=5BD$ where $BD=\sqrt {10^2-AD^2}$. Now,I am stuck. Can someone help? $\endgroup$ – learner Apr 16 '15 at 6:42

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