Complex Numbers
Complex Number
nta_pyq_2025_jan
Grade 11
Question:
Let $O$ be the origin, the point $A$ be $z_{1}=\sqrt{3}+2\sqrt{2}\,i$, the point $B(z_{2})$ be such that $\sqrt{3}|z_{2}|=|z_{1}|$ and $\arg(z_{2})=\arg(z_{1})+\dfrac{\pi}{6}$. Then:
area of triangle $ABO$ is $\dfrac{11}{\sqrt{3}}$
$ABO$ is an obtuse angled isosceles triangle
area of triangle $ABO$ is $\dfrac{11}{4}$
$ABO$ is a scalene triangle
Step-by-Step Solution
Key Concept: Compute $OA, OB$ from the moduli and apply the Law of Cosines at angle $\angle AOB=\dfrac{\pi}{6}$. Compare side lengths to classify the triangle.
$|z_{1}|=\sqrt{3+8}=\sqrt{11}$, so $|z_{2}|=\dfrac{\sqrt{11}}{\sqrt{3}}$. Let $OA=\sqrt{11},\ OB=\sqrt{11/3}$, and $\angle AOB=\dfrac{\pi}{6}$.
Law of Cosines:
$$AB^{2}=OA^{2}+OB^{2}-2\,OA\cdot OB\cos\frac{\pi}{6} = 11+\frac{11}{3}-2\sqrt{11}\cdot\sqrt{\frac{11}{3}}\cdot\frac{\sqrt{3}}{2} = 11+\frac{11}{3}-11 = \frac{11}{3}.$$
So $AB=\sqrt{11/3}=OB$ — the triangle is isosceles with $OB=AB$.
Since $OB=AB$, the angles \emph{opposite} those sides are equal: $\angle A=\angle O=\dfrac{\pi}{6}$. Therefore
$$\angle B = \pi-\frac{\pi}{6}-\frac{\pi}{6}=\frac{2\pi}{3}\ (>\frac{\pi}{2}).$$
Hence $\triangle ABO$ is an \textbf{obtuse-angled isosceles} triangle.
Correct Answer: 2