The area (in sq units) of the triangle whose | vedclass

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(B) Let the vertices of the triangle be $O(0,0)$,$A(z)$,and $B(z e^{i \alpha})$.The distance $OA = |z - 0| = |z|$.The distance $OB = |z e^{i \alpha} -

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$\frac{1}{2}|z|^2 \sin \alpha$

Vedclass · Answer

(B) Let the vertices of the triangle be $O(0,0)$,$A(z)$,and $B(z e^{i \alpha})$.The distance $OA = |z - 0| = |z|$.The distance $OB = |z e^{i \alpha} - 0| = |z| |e^{i \alpha}| = |z| 	imes 1 = |z|$.The angle between the vectors $OA$ and $OB$ is the argument of $\frac{z e^{i \alpha}}{z} = e^{i \alpha}$,which is $\alpha$.The area of a triangle with two sides $a$ and $b$ and included angle $	heta$ is given by $\frac{1}{2} ab \sin 	heta$.Therefore,the area of the triangle is $\frac{1}{2} 	imes OA 	imes OB 	imes \sin \alpha = \frac{1}{2} |z| |z| \sin \alpha = \frac{1}{2} |z|^2 \sin \alpha$.

The area (in sq units) of the triangle whose vertices are the points represented by the complex numbers $0, z$,and $z e^{i \alpha}$ $(0 < \alpha < \pi)$ is:

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