2 boats left the harbour at the same time. One travelled at 10km/h on a course of 47 degrees. from the north. the other travelled 8km/h on a course of 79 degrees. how far apart were the boats after 45 minutes to th nearest tenth of a kilometre.
the angle between their courses is 32°
using the law of cosines, the distance between the ships is thus
d^2 = (10*3/4)^2 + (8*3/4)^2 - 2*(10*3/4)(8*3/4) cos 32°
d^2 = 15.93
d = 4.0 km
To find the distance between the two boats after 45 minutes, we need to calculate the distance each boat has traveled in that time.
Let's start with the boat traveling at 10 km/h on a course of 47 degrees from the north.
Distance traveled by boat 1 in 45 minutes = (10 km/h) * (45/60) hours
= 7.5 km
Now, let's calculate the distance traveled by the boat traveling at 8 km/h on a course of 79 degrees.
Distance traveled by boat 2 in 45 minutes = (8 km/h) * (45/60) hours
= 6 km
Next, we will use the law of cosines to find the distance between the two boats.
cos(d) = cos(90 - θ1) * cos(90 - θ2) + sin(90 - θ1) * sin(90 - θ2) * cos(γ)
Where:
d is the distance between the two boats
θ1 and θ2 are the angles of courses of the two boats from the north
γ is the difference between the two angles of courses
In this case, θ1 = 47 degrees, θ2 = 79 degrees, and γ = θ2 - θ1 = 79 - 47 = 32 degrees.
cos(d) = cos(90 - 47) * cos(90 - 79) + sin(90 - 47) * sin(90 -79) * cos(32)
= cos(43) * cos(11) + sin(43) * sin(11) * cos(32)
= 0.683 + 0.891 * 0.192
= 0.683 + 0.170
= 0.853
Now, we can calculate the distance between the two boats using the inverse cosine function:
d = acos(0.853)
≈ 30.26 degrees (rounded to the nearest tenth)
Therefore, the boats are approximately 30.3 km apart after 45 minutes.
To find the distance between the two boats after 45 minutes, we can use the concept of vectors. Let's break down the problem into smaller steps:
1. Convert 45 minutes to hours. Since there are 60 minutes in an hour, 45 minutes is equal to 45/60 = 0.75 hours.
2. Use the formula v = d/t to find the distance traveled by each boat. The first boat traveled at 10 km/h for 0.75 hours, so the distance it traveled is 10 km/h * 0.75 h = 7.5 km. The second boat traveled at 8 km/h for 0.75 hours, so the distance it traveled is 8 km/h * 0.75 h = 6 km.
3. Convert the courses of the boats to angles in radians. To convert from degrees to radians, use the formula radians = degrees * (pi/180). The course of the first boat is 47 degrees, so the angle in radians is 47 * (pi/180) = 0.8203 radians. The course of the second boat is 79 degrees, so the angle in radians is 79 * (pi/180) = 1.3788 radians.
4. Calculate the horizontal and vertical components of the distance traveled by each boat using trigonometry. The horizontal component is given by distance * cos(angle), and the vertical component is given by distance * sin(angle).
For the first boat:
- Horizontal component = 7.5 km * cos(0.8203 radians) = 7.5 km * 0.6734 = 5.0505 km
- Vertical component = 7.5 km * sin(0.8203 radians) = 7.5 km * 0.7390 = 5.5426 km
For the second boat:
- Horizontal component = 6 km * cos(1.3788 radians) = 6 km * 0.1752 = 1.0511 km
- Vertical component = 6 km * sin(1.3788 radians) = 6 km * 1.3595 = 8.1570 km
5. Subtract the horizontal and vertical components of the two boats to find the distance between them. Use the distance formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the two boats.
- Horizontal distance = 5.0505 km - 1.0511 km = 3.9994 km
- Vertical distance = 5.5426 km - 8.1570 km = -2.6144 km
Since the vertical distance is negative, it means that the second boat is below the first boat. To find the actual distance between them, we can use Pythagoras' theorem: distance = sqrt(horizontal distance^2 + vertical distance^2) = sqrt((3.9994 km)^2 + (-2.6144 km)^2) = sqrt(15.995 km^2 + 6.8334 km^2) = sqrt(22.8284 km^2) = 4.77 km (rounded to the nearest tenth).
Therefore, the two boats are approximately 4.8 km apart after 45 minutes.