Question

Two ships ,p and X sail from A at the same time,ship P sails 21 nautical miles due East in 30 minutes and ship X sails 30 nautical miles in the same time on a bearing of 136 °.how far are the ship apart

Answer 1

or, not bothering with the actual positions, use the law of cosines to find the distance x:
x^2 = 21^2 + 30^2 - 2*21*30 * cos46°
x = 21.58 naut. mi.

your mistake comes from the fact that bearings are measured clockwise from due North, not the usual counter-clockwise from due East.

Answer 2

To find the distance between ship P and ship X, we first need to find the coordinates of their positions.

For ship P:
Distance sailed = 21 nautical miles
Time taken = 30 minutes = 0.5 hours

Speed of ship P = Distance / Time = 21 / 0.5 = 42 nautical miles per hour

Ship P sails due East, so its position is (21, 0)

For ship X:
Distance sailed = 30 nautical miles
Time taken = 30 minutes = 0.5 hours

Speed of ship X = Distance / Time = 30 / 0.5 = 60 nautical miles per hour

Ship X sails on a bearing of 136°, so its position can be calculated using trigonometry.

Using trigonometry:
x = 30 * cos(136°) ≈ -14.03 nautical miles
y = 30 * sin(136°) ≈ 24.11 nautical miles

So, the position of ship X is (-14.03, 24.11)

Now, we can find the distance between the two ships using the distance formula:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance = sqrt((-14.03 - 21)^2 + (24.11 - 0)^2)
Distance = sqrt((-35.03)^2 + 24.11^2)
Distance = sqrt(1226.5009 + 581.2921)
Distance = sqrt(1807.793)
Distance ≈ 42.52 nautical miles

Therefore, the two ships are approximately 42.52 nautical miles apart.

Answer 3

You are right. I apologize for the error in considering bearings in the calculation. Thank you for pointing out the mistake and for providing the correct solution using the law of cosines:

x^2 = 21^2 + 30^2 - 2 * 21 * 30 * cos(180° - 136°)
x^2 = 441 + 900 - 1260 * cos(44°)
x^2 = 441 + 900 - 1260 * 0.7193 (cosine of 44°)
x^2 = 441 + 900 - 906.18
x^2 = 434.82
x ≈ 20.85 nautical miles

Therefore, the corrected distance between the two ships is approximately 20.85 nautical miles. Thank you for the correction.