Let O be the origin, the point A be z₁=√3+2 √2 i, the point B(z₂) be such that √3|z₂|=|z₁| and arg (z₂)=arg (z₁)+π/6. Then

Solution

Maths sir · 4 steps

You know two sides and the angle between them at O, so find the third side with the cosine rule.

1. Two sides at O. OA = |z₁| = √(3 + 8) = √11. Then OB = |z₂| = √11/√3, and the angle between them is ∠ AOB = π/6, because arg(z₂) exceeds arg(z₁) by exactly that much.

   AB² = 11 + 11/3 - 2√11·√11/√3 cos π/6 = 11 + 11/3 - 11 = 11/3

2. Shape of the triangle. So AB = √11/√3 = OB, and the triangle is isosceles. Equal sides face equal angles, so ∠ A = ∠ O = 30°, which leaves ∠ B = 120°. That angle is obtuse.
3. Check the area. Area = 1/2 · OA · OB · sin 30° = 1/2 · 11/√3 · 1/2 = 11/4√3. This matches neither (a) nor (b).
4. So. ABO is an obtuse-angled isosceles triangle, which is option (d).

The idea to keep. Modulus gives you the length from the origin, and a difference of arguments gives you the angle at the origin. With those three facts, the triangle becomes a plain cosine-rule problem.

Answer from the NTA final answer key. Solution by PrepWiser.