Question

If sin^–1$\left\{ \frac{1}{2i}(z - 3) \right\}$be the angle of a triangle and if z = x + iy, then –

• A. x = 1, y = 3
• B. x = 3, 0 < y £ 2
• C. x = 1, y = 2
• D. x + y = 1

Answer 1

Answer: (B)

x = 3, 0 < y £ 2

Answer 2

Sol. Since sin^–1 $\left\{ \frac{1}{2i}(z - 3) \right\}$= q, say, is the angle of triangle, we must have $\frac{1}{2i}$ (z – 3) real or – $\frac{i}{2}$ (x + iy – 3)

or –$\frac{1}{2}$ [((x – 3) i – y)] = real

\ x – 3 = 0 or x = 3 … (1)

and $\frac{y}{2}$ is real or sin^–1 $\frac{y}{2}$ = q is the angle of a triangle

\ $\frac{y}{2}$ = sin q where 0 < sin q £ 1 as q cannot be

–ive or zero, 0 < $\frac{y}{2}$ £ 1 or 0 < y £ 2 and x = 3.