Initially, two ships and a port form an equilateral triangle with a side length of 24 km.
Given triangle ABC is equilateral with side 24 km, the coordinates of \( C \) are: \[ C = \left(12\sqrt{3}, 12\right) \] (Triangle positioned symmetrically relative to base AB.)
Since triangle \( APQ \) is right-angled at \( P \) and \( AP = 16 \), the following must hold: \[ PQ = AQ = 16 \]
The slower ship travels 8 km while the faster ship travels 24 km, indicating a speed ratio of \( 1:3 \). This implies both ships commence movement simultaneously, and the faster ship reaches the port precisely when the slower ship has covered 8 km.
At this point, triangle \( APQ \) is right-angled at \( P \), and \( AQ = 16 \). Therefore: \[ \text{Distance from Q to port A} = AQ = 16 \text{ km} \] As the triangle is right-angled at \( P \), the side \( PQ \) is perpendicular. Using triangle properties: \[ PQ = 12 \text{ km} \] This is the perpendicular distance from Q to the base AP.
✅ Final Answer: The distance between the remaining ship and the port is 12 km.
In the figure O is the centre of the circle and A, B, C are points on the circle. AOB = 50^, BOC = 80^. 