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What is the area of the sector ?​

What is the area of the sector ?​-example-1

1 Answer

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They forget to say "not to scale". I'm guessing this is trig because I don't see another way to do it.

Let's consider x=chord PQ first. By the Law of Cosines


x^2 = 3^2 + 5^2 - 2(3)(5) \cos 81^\circ = 34 - 30\cos 81^\circ

We have an isosceles triangle formed by two radii of 9 cm and x=PQ. By the Law of Cosines again,


x^2 = 9^2 + 9^2 - 2(9)(9) \cos B


x^2 = 162 - 162 \cos B


\cos B = (162 - x^2)/(162) = 1 - (x^2)/(162)


\cos B = 1 - (34 - 30\cos 81^\circ)/(162)


B = \arccos \left(1 - (34 - 30\cos 81^\circ )/(162) \right)

The area of the circle is the fraction given by the angle,


A = (B)/(360^\circ) \pi r ^2


A \approx ( \arccos \left(1 - (34 - 30\cos 81^\circ)/(162) \right) )/(360) (3.142)9^2


A\approx 24.7474332960707

Answer: 24.7 sq cm

User Rocketsarefast
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