Find the area of the circle circumscribed to a $\triangle ABC$, below.

Question

Calculate the area of the circle circumscribed to a $\triangle ABC$, such that $AB=BC$, $AC=6$ and measure of angle $ABC =37^o$ (S:$25 \pi$)

Is it possible to solve without using a calculator?

Answer 1

Let your $H$ be placed at $(0,0)$ and your $C$ be placed at $(3,0)$.

Using the 'Angle at center theorem' which states that $2\angle ABC=\angle AOC$, we know then that point $O$ is located at $\left(0,\frac{3}{\tan(37°)}\right)$.

This gives a distance $OC=R=\sqrt{9+\left(\frac{3}{\tan(37°)}\right)^2}=\frac{3}{\sin(37°)}$ which is roughly $R\approx 4.98492$.

The circle's area, with $a=R^2π$, is then
$a\approx 24,85π$.

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As for figuring this out without a calculator, best would probably be to use the Taylor series for $(\sin)$ in radians:$$\sin(x) = x-\frac{x^3}{3!}+\frac{x^5}{5!}- ...$$ This is doable - as well as a strong point towards using a calculator.

Answer 2

One can get the circumradius at once, using the (extended) law of sines (see here):

$$\frac{a}{\sin(A)}=2R$$

one gets :

$$R=\frac{a}{2 \sin(A)}$$