A space vehicle has burnout speed 13.5 km/s at 5° above the local horizon around Earth; the source describes its orbit as circular. Find the tangential component in km/s, to 2 decimals.
The tangential direction is along the local horizon. Resolve the given velocity: $V_t=V\cos5^\circ=13.5\cos5^\circ=13.4486$ km/s, or 13.45. Strictly, an exactly circular orbit has no radial velocity and zero flight-path angle; the printed 5° condition is used for the requested component rather than enforcing the inconsistent circular label.
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