Area of a cyclic triangle

Slightly different approach:

Let O be the center of the circle. Then since $OA = OB = AB = 1, \triangle OAB$ is equilateral and $\angle AOB = 60^\circ$ . Next, $\angle ACB = \frac{1}{2} \angle AOB = 30^\circ$ as they both subtend arc $AB$ ; this makes $\angle ABC = 100^\circ$ . Using sine law,

$\begin{aligned} \dfrac{\sin 100^\circ}{AC} &= \dfrac{\sin 30^\circ}{1} \\ \\ \dfrac{\sin 100^\circ}{AC} &= \dfrac{1}{2} \\ \\ AC &= 2 \sin 100^\circ \end{aligned}$

Finally, the area $A$ of $\triangle ABC$ is

$A = \dfrac{1}{2} (AB)(AC) sin \angle BAC = \dfrac{1}{2} (1)(2 \sin 100^\circ) \sin 50^\circ \approx \boxed{0.754406506735}$

• Let O be the center of the circle.

• OA=OB=AB as they are all radii $\therefore \triangle OAB$ is equilateral.

• $\angle OAC\quad =\quad 60-\angle CAB\quad =\quad 10deg$

• Using the circle theorem "The angle subtended at the centre is twice the angle subtended at the circumference from the same points". $\angle COB\quad =\quad 2\quad \times \quad \angle BAC\quad =\quad 100deg$

• $\triangle OBC$ is issoceles as two sides are radii. Thus, $\angle OBC\quad =\quad \angle OCB\quad =\quad \frac { 1 }{ 2 } (180-\angle COB)\quad =\quad 40\quad$

• $\angle ACB\quad =\quad 40-10\quad =\quad 30deg$

• From here, use the sine rule, $\frac { a }{ \sin { A } } =\frac { b }{ \sin { B } } =\frac { c }{ \sin { C } }$ , To work out length BC.

• $BC\quad =\quad \frac { 1\quad \times \quad \sin { 50 } }{ \sin { 30 } } \quad =\quad 1.532\quad (3\quad d.p)$

• Now, use $\frac { 1 }{ 2 } ab\sin { C }$ to work out the area.

• $\frac { 1 }{ 2 } \quad \times \quad 1.532...\quad \times \quad 1\quad \times \quad (180-50-30)\quad =\quad 0.754406506{ m }^{ 2 }$