Question

A sector with an area of \goldE{48\pi\,\text{cm}^2}48πcm

Answer 1

Answer:

$\theta=\dfrac{3\pi}{8} (in radians)$

Step-by-step explanation:

Area of a sector$=\dfrac{\theta}{2\pi}X\pi r^2$

Given: Area of a sector $=48\pi cm^2$

Radius of the circle =16cm

Therefore:

$48\pi cm^2=\dfrac{\theta}{2}X 16^2\\256\theta=96\pi\\\theta=\dfrac{96\pi}{256} \\\theta=\dfrac{3\pi}{8} (in radians)\\Therefore, the central angle \theta=\dfrac{3\pi}{8} (in radians)$