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Chapter 6: Problem 39
Two ships leave a harbor at the same time. One ship travels on a bearing of \(\mathrm{S} 12^{\circ} \mathrm{W}\) at 14 miles per hour. The other ship travels on a bearing of \(\mathrm{N} 75^{\circ} \mathrm{E}\) at 10 miles per hour. How far apart will the ships be after three hours? Round to the nearest tenth of a mile.
Short answer, step by step solution, calculate the distance travelled by both ships., convert the bearings of the two ships to standard mathematical bearings., realize the problem can be solved using law of cosines., calculate the distance between the two ships., key concepts.
These are the key concepts you need to understand to accurately answer the question.
Bearings in Navigation
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Question: 5 Question 5 2 yachts leave from a harbour, H. Yacht A leaves on a bearing of 065° and sails for 25km. Yacht Bleaves on a bearing of 120° and sails until it is 30km from Yacht A. How far has Yacht B sailed? (7 marks) I
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Two boats leave a port at the same time, one going north, the other traveling south. The northbound boat travel 18 mph faster than the southbound boat. If the southbound boat is traveling at 52 mph, how long will it be before they are 1586 miles apart?
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A yacht sails from a harbour H to a point C, then to a point D as shown below. C is 50 kilometres due east of H.. D is on a bearing of 040° from C and is 79 kilometres from H. Calculate the size of angle CDH.
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SOLUTION: two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 3 mi/h faster then the southbound boat. after two hou
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Two yachts leave harbour H. Yacht A sails on a bearing of 072º for . 30 km and then stops. Yacht B sails on a bearing of 140º for 50 km and then stops. How far apart are the two yachts when they both have stopped? ... One yacht is at 72 degrees and the other one is at 140. This means the angle between the paths of the yachts is 68 degrees.
Microsoft Word - Credit Past Paper Questions by Topic.doc. 9. Trigonometry 2 - Sine, Cosine Rule, Area of Triangle. 1. Two yachts leave from harbour H. Yacht A sails on a bearing of 072° for 30 kilometres and stops. Yacht B sails on a bearing of 140° for 50 kilometres and stops.
2 yachts leave from a harbour, H. Yacht A leaves on a bearing of 065° and sails for 25km. Yacht B leaves on a bearing of 120° and sails until it is 30km from Yacht A. Calculate how far Yacht B has sailed. Gauth AI Solution. Super Gauth AI. Yacht B has sailed approximately 35.9 km.
VIDEO ANSWER: I'm hi. The two yards are leaving from the same harbor. The first year is living on a bearing of 65 degrees, so let's take this as the harbor we have. 65 degrees bearing is what it's going to be. The coordinate lines are X,X. And axis
Yacht A sails on a bearing of 072° for 32 miles, and Yacht B sails on a bearing of 140° for 50 miles. We need to use trigonometry to solve this problem, treating the problem as one of finding the third side of a triangle when two sides and the included angle are known. First, we calculate the angle between the paths of the two yachts.
Question 5.2 yachts leave from a harbour, H.Yacht A leaves on a bearing of 065° and sails for 25km.Yacht B leaves on a bearing of 120° and sails until it is 30km from Yacht A.C Calculate how far Yacht B has sailed.[7 marks]
2 yachts leave from a harbour, H. Yacht A leaves on a bearing of 0650 and sails for 25km. Yacht B leaves on a bearing of 1200 and sails until it is 30km from...
In the 2nd part of your problem. Work out the angle between the two headings. The relative velocity will be the third side of the triangle of velocities, the magnitude of which can be obtained by the cos law: c 2 = a 2 + b 2 - 2abcosθ. The relative velocity times the time will be the relative displacement of the two boats.
From the above diagram 13∘ 13 ∘ is the angle travelled towards west and 75∘ 75 ∘ is the angle travelled towards East. 220 220 is the distance travelled towards west in 22 22 hours. 176 176 is the distance travelled towards East in 22 22 hours. Now we need to find x x and you can use Law of Cosines. x2 = (220)2 + (176)2 − 2(220)(176 ...
Two ships leave a harbor at the same time. One ship travels on a bearing of \(\mathrm{S} 12^{\circ} \mathrm{W}\) at 14 miles per hour. ... In our exercise, bearings were used to set the courses of two ships leaving a harbor. Bearings are given in degrees, with North as 0° or 360°, East as 90°, South as 180°, and West as 270°. The bearings ...
9. Trigonometry 2 - Sine, Cosine Rule, Area of Triangle 1. Two yachts leave from harbour H. Yacht A sails on a bearing of 072° for 30 kilometres and stops. Yacht B sails on a bearing of 140° for 50 kilometres and stops. How far apart are the two yachts when they have both stopped? Do not use a scale drawing. 4 RE 2.
IC. A yacht sails from a harbour H to a point C, then to a point D as shown below. North D 79 km North North H C 50 km C is 50 kilometres due east of H. D is on a bearing of 040 ° from C and is 79 kilometres from H. a Calculate the size of angle CDH. b Hence, calculate the bearing on which the yacht must sail to return directly to the harbour.
Lisa B. asked • 02/16/13 two boats leave the harbor at 9:00 am. Boat A sails north at 20 km/hr. Boat B sails west at 15km/hr. 158, How far apart are the two boats at noon?
( Stéei) 2022-23) Two yachts leave from harbour H. Yacht A sails on a bearing of 072° for 30 kilometres and stops. Yacht B sails on a bearing of 140° for 50 kilometres and stops. figures. * (1 Point) How far apart are the two yachts when they have both stopped? Answer to 3 significant N Enter yo
Question: 5 Question 5 2 yachts leave from a harbour, H. Yacht A leaves on a bearing of 065° and sails for 25km. Yacht Bleaves on a bearing of 120° and sails until it is 30km from Yacht A. How far has Yacht B sailed? (7 marks) I. Show transcribed image text. Here's the best way to solve it.
The distance is changing at sqrt(1476)/2 knots per hour. Let the distance between the two boats be d and the number of hours they've been travelling be h. By the pythagorean theorem, we have: (15h)^2 + (12h)^2 = d^2 225h^2 + 144h^2 = d^2 369h^2 = d^2 We now differentiate this with respect to time. 738h = 2d((dd)/dt) The next step is finding how far apart the two boats are after two hours. In ...
Two yachts leave from harbour H. Yacht A sails on a bearing of 072° for 30 kilometres and stops. Yacht B sails on a bearing of 140° for 50 kilometres and stops.How far apart are the two yachts when they have both stopped? 7. The area of the triangle shown is 36 cm2. Show that Sin R= 3/4. 8. The path in the diagram opposite runs parallel to ...
Northbound boat speed is 52 + 18 = 70mph. Since distance is speed x time let time = t Then: 52t + 70t = 1586 solving for t 122t = 1586 => t = 13 t = 13 hours Check: Southbound (13)(52) = 676 Northbound (13)(70) = 910 676 + 910 = 1586 ... Two boats leave a port at the same time, one going north, the other traveling south. The northbound boat ...
In this problem, we have the position of the harbour here: the 2 ships- let's be h, 2, and let this be 1 such that the distance is 402 and we need to figure out the distance as 1 s 2, given that the simbonis 135 degrees and 40 Minutes, thinknote that if there are in 60 arc minutes 1 degree, 60 arc minutes is equivalent to 1 degree, which means 40 arc minutes will be equal to 40 by 60 degrees ...
D C is 50 kilometres due east of H. D is on a bearing of 040 ° from C and is 79 kilometres from ... 5 Stéei 2022-23 Two yachts leave from harbour H. Yacht A sails on a bearing of 072 ° for 30 kilom...
how far apart are the two ships? Two ships leave a harbor at the same time, traveling on courses that have an angle of 100 degrees between them. If the first ship travels at 20 miles per hour and the second ship travels at 25 miles per hour, how far apart are the two ships after 3.1 hours? Follow • 1. Add comment.
Question 564723: two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 3 mi/h faster then the southbound boat. after two hours the boats are 30 mi apart. find the speed of the southbound boat Answer by mananth(16942) (Show Source):