Maths problem:how to do (c), i found out angle AOB=68° and radius is 4.82 cm. Thanks?

(a) From △AOB :
∵ radius of semicircle, r = OA = OB, 
∴ △AOB is an isos.△　　
=> ∠ABO = 56⁰ 　　(base ∠s, isos.△) 
So, ∠AOB = 180⁰-2(56⁰) = 68⁰

(b) From △ABC : ∠C=90-56=34⁰
From △BOC :
r² = 8²+r²-2(8r) cos 34⁰ 　　
0 = 64-16r cos 34⁰
r cos 34⁰ = 4
∴ r = 4/cos 34⁰ = 4.82 cm (to 2 dec.pl.)

# (a),(b) 都答對 !! *\(^o^)/*

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(c) area of △AOB=½ r*r sin 68⁰=½(4.82)²sin68⁰
area of sector AOB=πr²(68⁰/360⁰)=π(4.82)²(68⁰/360⁰)
∴ area of shaded part
= π(4.82)²(68⁰/360⁰) - ½(4.82)²sin68⁰
= (4.82)² [(68⁰/360⁰)π - (sin68⁰)/2]
= 3.02 (to 2 dec.pl.)