A sector with an area of \goldE{48\pi\,\text{cm}^2}48πcm
2
start color #a75a05, 48, pi, start text, c, m, end text, squared, end color #a75a05 has a radius of \maroonD{16\,\text{cm}}16cmstart color #ca337c, 16, start text, c, m, end text, end color #ca337c.



What is the central angle measure of the sector in radians?

Answer :

Answer:

[tex]\theta=\dfrac{3\pi}{8} $ (in radians)[/tex]

Step-by-step explanation:

Area of a sector[tex]=\dfrac{\theta}{2\pi}X\pi r^2[/tex]

Given: Area of a sector [tex]=48\pi cm^2[/tex]

Radius of the circle =16cm

Therefore:

[tex]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)[/tex]

1368754

Answer:

3pi/8

Step-by-step explanation:

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