Diagonals of pentagon

Calculate the diagonal length of the regular pentagon:
a) inscribed in a circle of radius 12dm;
b) a circumscribed circle with a radius of 12dm.

Correct result:

u1 =  22.825 dm
u2 =  28.214 dm


r1=12 cm A=360/(2 5)=36   sinA=a1/2:r1  a1=2 r1 sinA=2 r1 sin36 =2 12 sin36 =2 12 0.587785=14.10685  u1=a1/2 (1+5)=14.1068/2 (1+5)=22.825 dmr_{1}=12 \ \text{cm} \ \\ A=360/ (2 \cdot \ 5)=36 \ ^\circ \ \\ \ \\ \sin A=a_{1}/2 : r_{1} \ \\ \ \\ a_{1}=2 \cdot \ r_{1} \cdot \ \sin A ^\circ =2 \cdot \ r_{1} \cdot \ \sin 36^\circ \ =2 \cdot \ 12 \cdot \ \sin 36^\circ \ =2 \cdot \ 12 \cdot \ 0.587785=14.10685 \ \\ \ \\ u_{1}=a_{1}/2 \cdot \ (1+\sqrt{ 5 })=14.1068/2 \cdot \ (1+\sqrt{ 5 })=22.825 \ \text{dm}
r2=12 cm  tanA=a2/2:r2  a2=2 r2 tanA=2 r2 tan36 =2 12 tan36 =2 12 0.726543=17.43702  u2=a2/2 (1+5)=17.437/2 (1+5)=28.214 dmr_{2}=12 \ \text{cm} \ \\ \ \\ \tan A=a_{2}/2 : r_{2} \ \\ \ \\ a_{2}=2 \cdot \ r_{2} \cdot \ \tan A ^\circ =2 \cdot \ r_{2} \cdot \ \tan 36^\circ \ =2 \cdot \ 12 \cdot \ \tan 36^\circ \ =2 \cdot \ 12 \cdot \ 0.726543=17.43702 \ \\ \ \\ u_{2}=a_{2}/2 \cdot \ (1+\sqrt{ 5 })=17.437/2 \cdot \ (1+\sqrt{ 5 })=28.214 \ \text{dm}

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