Circular sector

I have a circular sector with a length 15 cm with an unknown central angle. It is inscribed by a circle with radius 5 cm. What is the central angle alpha in the circular sector?

Result

A =  0 °

Solution:

r(2+A)=15 r2=5 cm cos⁡A/2=r2/(r−r2) o2=2π⋅ r2=2⋅ 3.1416⋅ 5≐31.4159 cm  =2π5=31.415 cm>15 cm o1<o2 A=P=0=0∘=0∘r(2+A) = 15 \ \\ r_{ 2 } = 5 \ cm \ \\ \cos A/2 = r_{ 2 } / (r- r_{ 2 }) \ \\ o_{ 2 } = 2 \pi \cdot \ r_{ 2 } = 2 \cdot \ 3.1416 \cdot \ 5 \doteq 31.4159 \ cm \ \\ \ \\ = 2\pi 5 = 31.415 \ cm > 15 \ cm \ \\ o_{ 1 } < o_{ 2 } \ \\ A=P = 0 = 0 ^\circ = 0^\circ

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