SOLUTION: Two ships P and Q leaves a point at the same tine . P sails at 10km/hr on a bearing of 030 degree and Q sails at 12km/hr on a bearing of 300 degree calculate their distances apart
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Question 335832: Two ships P and Q leaves a point at the same tine . P sails at 10km/hr on a bearing of 030 degree and Q sails at 12km/hr on a bearing of 300 degree calculate their distances apart and the bearing of P from Q after 2 hours
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
P sails at 10 km/hr on a bearing of 30 degrees.
Q sails at 12 km/hr on a bearing of 60 degrees.
Plot their vectors on a grid starting from the center, and you will see that their combined vectors form a right triangle that has one leg with a length of 10 kilometers per hour, and another leg with a length of 12 kilometers per hour.
If you mark the center of the grid as A, then P goes from A to B forming a 30 degree angles with the horizontal axis, and Q goes from A to C forming a negative 60 degrees with the horizontal axis.
The triangle formed is ABC, where the 90 degree angle is at A (30 degrees plus 60 degrees = 90 degrees).
Side a is opposite angle A and is formed by the line segment BC.
Side b is opposite angle B and is formed by the line segment AC.
Side c is opposite angle C and is formed by he line segment AB.
Since the 90 degree angle is at angle A, then the hypotenuse of this right triangle is side a which is formed by the line segment BC.
Since this is a right triangle, you can find the hypotenuse of this triangle by using the Pythagorean theorem of a^2 + b^2 = c^2.
You get 10^2 + 12^2 = 100 + 144 = 244.
Square root of 244 is equal to 15.62049935.
This means that the points B and C are increasing the distance between them at a rate of 15.62049935 kilometers per hour.
In 2 hours, this means that the ships are 31.2409987 miles apart.
In 4 hours, the ships will be 62.48199741 miles apart, etc.
If the angle formed by them was not 90 degrees, then you could have found the third side using the Law of Cosines formula of:
a^2 = b^2 + c^2 - 2*b*c*cos(A).
With this formula, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.
As long as their velocity remains the same, the distance between them will continue to increase by that velocity multiplied by the time in hours.
You can see this better by creating a distance vector rather than a velocity vector.
The distance vector depends on time.
In 1 hour, P travels a distance of 10 kilometers at a bearing of 30 degrees and Q travels a distance of 12 kilometers at a bearing of 300 degrees (-60 degrees).
In 2 hours, P travels a distance of 20 kilometers at a bearing of 30 degrees and Q travels a distance of 24 kilometers as a bearing of 300 degrees.
The triangle formed for the 2 hour distances is still a right triangle labeled ABC where A is the origin of both boats, and B is the point that P has reached, and C is the point that Q has reached. Triangle ABC is therefore labeled with a 90 degree at angle A, and side AB = 20 kilometers, and side AC = 24 kilometers.
Since side AB is opposite angle C, then side AB is called side c.
Since side AC is opposite angle B, then side AC is called side b.
The side opposite angle A is called side a which is formed by the line segment BC.
You want to find the length of side a because it is the distance between the 2 boats.
Using the right triangle formula and since side a is the hypotenuse of the right triangle formed, you get a^2 = b^2 + c^2 which becomes a^2 = 20^2 + 24^2 = 31.2409987 kilometers.
This is exactly what we derived above.
Using the Law of Cosines, the formula becomes:
a^2 = b^2 + c^2 - 2*b*c*cos(C).
this becomes:
a^2 = 20^2 + 24^2 - 2*20*24*cos(90).
Since cos(90) = 0, this formula becomes:
a^2 = 20^2 + 24^2 which is, once again, exactly what we derived above.
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