Triangle Area Formula In , given , , and . Find the area of .

Trigonometry Solution – Problem 25: Find the area of

Triangle Area Formula

In $\triangle ABC$ , given $B = 60°$ , $\sin A = 2\sin C$ , and $b = 2\sqrt{3}$ . Find the area of $\triangle ABC$ .

From $\sin A = 2\sin C$ , by Law of Sines: $a = 2c$ .

Using Law of Cosines:

$b^2 = a^2 + c^2 - 2ac\cos B$

$12 = 4c^2 + c^2 - 2 \cdot 2c \cdot c \cdot \frac{1}{2}$

$12 = 5c^2 - 2c^2 = 3c^2$

$c = 2, \quad a = 4$

Area:

$S = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$

$2\sqrt{3}$

Choose the correct theorem (2 pts): Law of Sines / Law of Cosines / area formula / circumradius relation as appropriate.
Set up equations correctly (2 pts): substitute given data and write a solvable system.
Solve and (if needed) reject extraneous cases (2 pts): handle SSA ambiguity or inequality constraints if present.
Final answer (1 pt): provide the requested length/angle/area in the required form.