∫∠BDC = 90 degrees.

∴∠CBD=90 degrees -60 degrees =30 degrees

∫∠CBD = 30 degrees

∴CD= 1/2BC (reason: in a right triangle, the right side facing an angle of 30 degrees is half the length of the hypotenuse)

BC=20 nautical miles, so CD= 10 nautical miles.

Pythagorean theorem can be found: BD = BC 2-CD 2 = 20 2-10 2 = 400-200 = 200 nautical miles.

So BD= 200 nautical miles under the square root.

Also know: ∠BAC=90 degrees -60 degrees =30 degrees.

It can also be seen that AD=2BD.

Because BD= 200 nautical miles under the arithmetic square root

∴AD=2 times 200 nautical miles under the arithmetic square root.

∫CD = 10 nautical mile (previously proved)

So AC=2 times 200 nautical miles-10 nautical miles under the arithmetic square root.

∴ The time for the ship to reach point C is: (2 times the arithmetic square root at 200 nautical miles-10 nautical miles) ∴10.

Note: The ship is sailing at the speed of 10 knots.

What is found here is the time from ∴a to C, as long as the ship passes by at 1 1: 30 in the morning and then arrives at C.

Just (because the root number can't be typed, the specific value is calculated by yourself): 1: 30.

(2): If the ship continues to sail eastward from point C, when will it reach point D, just south of island B?

What is required is: the length of CD 65438+the speed of 00 knots.

Let's understand: CD= 10 nautical mile, so it is exactly the line 10÷ 10= 1 hour.

∴ 165438+ 0: 30am 1 hour, and the required time is: 12: 30.

Plus the time spent on (1) is 2: 30.