The spacecraft

The spacecraft spotted a radar device at altitude angle alpha = 34 degrees 37 minutes and had a distance of u = 615km from Earth's observation point.
Calculate the distance d of the spacecraft from Earth at the moment of observation. Earth is considered a sphere with a radius of 6378km.

Correct result:

x =  368.383 km


α=34+3760=20776034.6167   u=615 km R=6378 km β=90+α=90+34.6167747760124.6167   (x+R)2=u2+R22 u R cosα  x=u2+R22 u R cosβR=u2+R22 u R cos124.61666666667 R=6152+637822 615 6378 cos124.61666666667 6378=6152+637822 615 6378 (0.568083)6378=368.383 kmα=34 +\dfrac{ 37 }{ 60 }=\dfrac{ 2077 }{ 60 } \doteq 34.6167 \ ^\circ \ \\ \ \\ u=615 \ \text{km} \ \\ R=6378 \ \text{km} \ \\ β=90 + α=90 + 34.6167 \doteq \dfrac{ 7477 }{ 60 } \doteq 124.6167 \ ^\circ \ \\ \ \\ (x+R)^2=u^2 + R^2 - 2 \cdot \ u \cdot \ R \cdot \ \cos α \ \\ \ \\ x=\sqrt{ u^2 + R^2 - 2 \cdot \ u \cdot \ R \cdot \ \cos β ^\circ } - R=\sqrt{ u^2 + R^2 - 2 \cdot \ u \cdot \ R \cdot \ \cos 124.61666666667^\circ \ } - R=\sqrt{ 615^2 + 6378^2 - 2 \cdot \ 615 \cdot \ 6378 \cdot \ \cos 124.61666666667^\circ \ } - 6378=\sqrt{ 615^2 + 6378^2 - 2 \cdot \ 615 \cdot \ 6378 \cdot \ (-0.568083) } - 6378=368.383 \ \text{km}

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