an asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend?

An asteroid has a diameter of 10km and is at a distance of 10^6km, what angle in degrees does it subtend ?

$$\tan{ ( \frac{ \Theta\ensurement{^{\circ}} } {2} ) } = \dfrac{ 5\ km } { 10^6\ km } = 5*10^{-7}\\\\ \frac{ \Theta\ensurement{^{\circ}} } {2} = 0.00002864789\ensurement{^{\circ}}\\\\ \Theta\ensurement{^{\circ}} =0.00005729578\ensurement{^{\circ}}\\\\ \Theta^{'}=\Theta\ensurement{^{\circ}} \times \dfrac{60^{'}} { 1\ensurement{^{\circ}} } = (0.00005729578 * 60)^{'} = 0.00343774677^{'}\\\\ \Theta^{"}=\Theta^{'}\times \dfrac{ 60^{"} } { 1^{'} } = (0.00343774677* 60)^{"} = 0.20626480625^{"}$$

60º