TITLE: Circle geometry, finding area of the minor segment
QUESTION [1 upvotes]: Just need a little bit of help with this question:
In the diagram below, the magnitude of the area of the minor segment cut off by chord AB is: (circle radius is 8, angle AOB is 105 degrees, don't know chord length.
I've tried manually solving it with the area of a segment formula (area of sector - triangle), and got $56\pi/3 - 32\sin(105)$, which is not the answer!
But I understand that area of the sector is $56\pi/3$, and triangle is $32\sin105$.
What is the question asking for?
Many thanks!

REPLY [1 votes]: Your calculation is correct. The given answer is wrong.
Minor segment area ( area difference between sector and triangle)
$$ =r^2 \theta/2 - r^2/2 \sin \theta $$
$$ =\dfrac {8^2 \cdot 105 \pi} {2 \cdot 180}-\dfrac{8^2\sin 105^{\circ}}{2}\text{
= what you got.}$$
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