Illustrate a scenario where two boats depart from a port at the same time. The first has a bearing of 135 degrees and is moving at a pace that would be equivalent to 15km/hr, giving it a brisk, early lead. The second, however, is sailing towards a bearing of 63 degrees with a slower pace that would mirror 2km/hr. After a span of what would equate to two hours, the second boat finds itself directly to the North of the first. Do not include any numerical values or calculations in the image. Allow the disparities of direction and speed to be evident.

Two boats leave a port at the sametime. The first travels at 15km/hr on a bearing of 135 degree. While the second travels at 2km/hr on a bearing of 63 degree . If after 2hours the second boat is directly North of the first boat, calculate their distance apart.

The distance should be gotten so as to know the bearing of the two boats. Since an average speed equals distance over time, therefore distance will be a.v(average speed)*time.

Distance=15*2
=30km for the first boat
And for the second boat,
Distance =20*2=40km
The angle between them is 72°.
Using cosine rule,
a^2=b^2+c^2-2bccosA
a^2=40^2+30^2-2*40*30cos72
a^2=1600+900-2400(0.309)
a^2=2500-741.6
a^2=1758.4
a=√1758.4
a=41.93~42km.

So the angle between them is 72°

They both travel for 2 hours, so the 1st went 30 km, and the 2nd went 4 km

This is just a cosine law question ....

x^2 = 30^2 + 4^2 - 2(4)(30)cos72
= 841.83..
x = appr 29.01 km

or, using the fact that the 2nd boat was directly north of the 1st, and noting parallel lines
4/sin45 = x/sin72
xsin45 = 4sin72
x = 4sin72/sin45 = 5.37 km

which is inconsistent with my other answer.
I think you have contradictory information.
Make a sketch and you will see that 2nd boat cannot be directly north of the 1st.

Check your typing or your question itself

TWO BOAT LEAVE A PORT AT THE SAME TIME THE FIRST TRAVELS AT 15KM /HR ON A BEARING 135 WHILE THE SECOND TRAVEL AT 20KM /HR ON A BEARING NORTH OF THE FIRST BOAT CALCULATE THEIR DISTANCE A PART

Why can't we use some rule for this question?

If we take the angle from at the port and the angle formed at the BOAT A, I believe we can use SINE RULE.

Well, it seems like these boats are playing a game of "who can get lost the fastest"! Let's see if we can help them out.

To solve this problem, we can break it down into two components: the horizontal (east-west) distance and the vertical (north-south) distance between the two boats.

First, let's find out how far the first boat has traveled in the horizontal direction. Using the formula distance = speed × time, we have:
Horizontal distance of the first boat = 15 km/hr × 2 hr = 30 km.

Now, let's find out how far the second boat has traveled in the vertical direction. Again, using the formula distance = speed × time, we have:
Vertical distance of the second boat = 2 km/hr × 2 hr = 4 km.

Since the second boat is directly north of the first boat, we only need to calculate the hypotenuse of the right triangle formed by the horizontal and vertical distances. Applying the Pythagorean theorem (a^2 + b^2 = c^2), where a and b are the horizontal and vertical distances respectively, we get:
Distance apart = √((30 km)^2 + (4 km)^2)
= √(900 km^2 + 16 km^2)
= √(916 km^2)
≈ 30.27 km.

So, after all those twists and turns, it turns out that the boats are approximately 30.27 km apart.

To calculate the distance between the two boats after 2 hours, we can use the concept of relative velocity. Relative velocity is the velocity of one object as observed from another object.

First, let's calculate the displacement of each boat after 2 hours. Displacement is the change in position of an object.

For the first boat:
Distance = Speed * Time
Distance = 15 km/hr * 2 hours
Distance = 30 km

The first boat has traveled 30 km in 2 hours on a bearing of 135 degrees.

For the second boat:
Distance = Speed * Time
Distance = 2 km/hr * 2 hours
Distance = 4 km

The second boat has traveled 4 km in 2 hours on a bearing of 63 degrees.

Now, we need to find the horizontal and vertical components of the distances traveled by each boat. We can use trigonometry to find these components.

For the first boat:
Horizontal distance = Distance * cos(bearing)
Horizontal distance = 30 km * cos(135 degrees)
Horizontal distance = 30 km * (-0.707)
Horizontal distance = -21.22 km (rounded to two decimal places)

Vertical distance = Distance * sin(bearing)
Vertical distance = 30 km * sin(135 degrees)
Vertical distance = 30 km * 0.707
Vertical distance = 21.22 km (rounded to two decimal places)

For the second boat:
Horizontal distance = Distance * cos(bearing)
Horizontal distance = 4 km * cos(63 degrees)
Horizontal distance = 4 km * 0.448
Horizontal distance = 1.79 km (rounded to two decimal places)

Vertical distance = Distance * sin(bearing)
Vertical distance = 4 km * sin(63 degrees)
Vertical distance = 4 km * 0.896
Vertical distance = 3.58 km (rounded to two decimal places)

Now, let's find the overall horizontal and vertical displacement between the two boats:

Overall horizontal displacement = Horizontal distance of first boat - Horizontal distance of second boat
Overall horizontal displacement = -21.22 km - 1.79 km
Overall horizontal displacement = -23.01 km (rounded to two decimal places)

Overall vertical displacement = Vertical distance of second boat - Vertical distance of first boat
Overall vertical displacement = 3.58 km - 21.22 km
Overall vertical displacement = -17.64 km (rounded to two decimal places)

Now, we can use the Pythagorean theorem to find the distance between the two boats:

Distance between the two boats = √(Overall horizontal displacement^2 + Overall vertical displacement^2)
Distance between the two boats = √((-23.01 km)^2 + (-17.64 km)^2)
Distance between the two boats = √(529.48 km^2 + 310.93 km^2)
Distance between the two boats = √(840.41 km^2)
Distance between the two boats ≈ 28.99 km (rounded to two decimal places)

Therefore, after 2 hours, the two boats are approximately 28.99 km apart.

no sketch