A) equilateral
B) right-angled
C) isosceles
D) None of these
Correct Answer: C
Solution :
\[z_{1}^{2}+z_{2}^{2}+2{{z}_{1}}{{z}_{2}}\cos \theta =0\] \[\Rightarrow \]\[{{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}^{2}}+2\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)\cos \theta +1=0\] \[\Rightarrow \]\[{{\left( \frac{{{z}_{1}}}{{{z}_{2}}}+\cos \theta \right)}^{2}}=-(1-{{\cos }^{2}}\theta )=-{{\sin }^{2}}\theta \] \[\Rightarrow \]\[\frac{{{z}_{1}}}{{{z}_{2}}}=-\cos \theta \pm i\sin \theta \] \[\Rightarrow \]\[\left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=\sqrt{{{(-\cos \theta )}^{2}}+{{(\sin \theta )}^{2}}}=1\] \[\Rightarrow \]\[|{{z}_{1}}|\,\,=\,\,|{{z}_{2}}|\] \[\Rightarrow \]\[|{{z}_{1}}-0|\,\,=\,\,|{{z}_{2}}-0|\] Thus, triangle with vertices\[0,\,\,{{z}_{1}},\,\,{{z}_{2}}\]is isosceles.
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