A ship on course of 253° T at 14 knots. At 2329 a lighthouse was observed bearing 282° T. At 2345 the same lighthouse bears 300° T. Find the ship’s distance off the second bearing and when abeam?
Given:
Course = 253°
Speed = 14 knots
First Bearing = 282°
Second Bearing = 300°
Find:
A.) Ship’s distance off the second bearing
B.) Ship’s distance when abeam
Solution:
1. Find first Angles A, B, C:
a.)Co = 253° T
Brg1 = 282° T
Angle A = 29° R
b.) Co = 253° T
Brg2 = 300° T
Angle B = 47° R
c.) Angle B = 47° R
Angle A = 29° R
Angle C = 18° R
2. Then solve for AB (Distance Run Between 1st & 2nd Observation)
AB = ( Time 2 – Time 1 ) x Speed
= ( 2345 – 2329 ) x 14 knots
= 16 minutes x 14 knots
AB = 3.73 nautical miles
3. Solve for Distance off at second bearing (BC): (By SINE Law)
BC = Sin A x AB
Sin C
BC = Sin 29° x 3.73
Sin 18°
BC = 5.9 nautical miles (Distance off at 2nd bearing)
4. Solve for Distance at Abeam (CD) : (By SOH – CAH – TOA)
CD = BC x Sin B
CD = 5.9 x Sin 47°
CD = 4.3 nautical miles ( Distance at Abeam)
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Given:
Course = 253°
Speed = 14 knots
First Bearing = 282°
Second Bearing = 300°
Find:
A.) Ship’s distance off the second bearing
B.) Ship’s distance when abeam
Solution:
1. Find first Angles A, B, C:
a.)Co = 253° T
Brg1 = 282° T
Angle A = 29° R
b.) Co = 253° T
Brg2 = 300° T
Angle B = 47° R
c.) Angle B = 47° R
Angle A = 29° R
Angle C = 18° R
2. Then solve for AB (Distance Run Between 1st & 2nd Observation)
AB = ( Time 2 – Time 1 ) x Speed
= ( 2345 – 2329 ) x 14 knots
= 16 minutes x 14 knots
AB = 3.73 nautical miles
3. Solve for Distance off at second bearing (BC): (By SINE Law)
BC = Sin A x AB
Sin C
BC = Sin 29° x 3.73
Sin 18°
BC = 5.9 nautical miles (Distance off at 2nd bearing)
4. Solve for Distance at Abeam (CD) : (By SOH – CAH – TOA)
CD = BC x Sin B
CD = 5.9 x Sin 47°
CD = 4.3 nautical miles ( Distance at Abeam)
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