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(B) The area of a triangle with vertices in polar coordinates $(r_1, 	heta_1)$,$(r_2, 	heta_2)$,and $(r_3, 	heta_3)$ is given by the formula: $	ex

Vedclass · Accepted Answer

$\frac{5\sqrt{3}}{4}$

Vedclass · Answer

(B) The area of a triangle with vertices in polar coordinates $(r_1, 	heta_1)$,$(r_2, 	heta_2)$,and $(r_3, 	heta_3)$ is given by the formula: $	ext{Area} = \frac{1}{2} |r_1 r_2 \sin(	heta_2 - 	heta_1) + r_2 r_3 \sin(	heta_3 - 	heta_2) + r_3 r_1 \sin(	heta_1 - 	heta_3)|$ Substituting the given values $(1, 0)$,$(2, \frac{\pi}{3})$,and $(3, \frac{2\pi}{3})$: $	ext{Area} = \frac{1}{2} |1 \cdot 2 \sin(\frac{\pi}{3} - 0) + 2 \cdot 3 \sin(\frac{2\pi}{3} - \frac{\pi}{3}) + 3 \cdot 1 \sin(0 - \frac{2\pi}{3})|$ $	ext{Area} = \frac{1}{2} |2 \sin(\frac{\pi}{3}) + 6 \sin(\frac{\pi}{3}) + 3 \sin(-\frac{2\pi}{3})|$ $	ext{Area} = \frac{1}{2} |2(\frac{\sqrt{3}}{2}) + 6(\frac{\sqrt{3}}{2}) + 3(-\frac{\sqrt{3}}{2})|$ $	ext{Area} = \frac{1}{2} |\sqrt{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{2}|$ $	ext{Area} = \frac{1}{2} |4\sqrt{3} - 1.5\sqrt{3}| = \frac{1}{2} |2.5\sqrt{3}| = \frac{5\sqrt{3}}{4} 	ext{ square units}$

The area (in square units) of the triangle formed by the points with polar coordinates $(1, 0)$,$(2, \frac{\pi}{3})$,and $(3, \frac{2\pi}{3})$ is:

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