Circular railway

The railway is to interconnect in a circular arc the points A, B, and C, whose distances are | AB | = 30 km, AC = 95 km, BC | = 70 km. How long will the track from A to C?

Result

b1 =  103.11 km

Solution:

c=30 b=95 a=70  s=a+b+c2=70+95+302=1952=97.5 S=s (sa) (sb) (sc)=97.5 (97.570) (97.595) (97.530)672.6522 km2  r=S/s=672.6522/97.56.899 km  R=a b c4 r s=70 95 304 6.899 97.574.1468 km  ΔRRb b2=R2+R22 R2 cosθ  θ=arccos(2 R2b22 R2)=arccos(2 74.146829522 74.14682)1.3906 rad  b1=θ R=1.3906 74.1468103.1095=103.11  km c = 30 \ \\ b = 95 \ \\ a = 70 \ \\ \ \\ s = \dfrac{ a+b+c }{ 2 } = \dfrac{ 70+95+30 }{ 2 } = \dfrac{ 195 }{ 2 } = 97.5 \ \\ S = \sqrt{ s \cdot \ (s-a) \cdot \ (s-b) \cdot \ (s-c) } = \sqrt{ 97.5 \cdot \ (97.5-70) \cdot \ (97.5-95) \cdot \ (97.5-30) } \doteq 672.6522 \ km^2 \ \\ \ \\ r = S/s = 672.6522/97.5 \doteq 6.899 \ km \ \\ \ \\ R = \dfrac{ a \cdot \ b \cdot \ c }{ 4 \cdot \ r \cdot \ s } = \dfrac{ 70 \cdot \ 95 \cdot \ 30 }{ 4 \cdot \ 6.899 \cdot \ 97.5 } \doteq 74.1468 \ km \ \\ \ \\ \Delta R-R-b \ \\ b^2 = R^2+R^2 - 2 \ R ^2 \ \cos θ \ \\ \ \\ θ = \arccos(\dfrac{ 2 \cdot \ R^2 - b^2 }{ 2 \cdot \ R^2 } ) = \arccos(\dfrac{ 2 \cdot \ 74.1468^2 - 95^2 }{ 2 \cdot \ 74.1468^2 } ) \doteq 1.3906 \ rad \ \\ \ \\ b_{ 1 } = θ \cdot \ R = 1.3906 \cdot \ 74.1468 \doteq 103.1095 = 103.11 \ \text{ km }

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