Please help with this question:
A plane travels a distance of 140 km due west of Vancouver International Airport. It then travels 60 km at a bearing of 340 degrees. How far from the Airport is the plane?
Thank you for all your help.
According to my sketch , we have a cosine law problem
x^2 = 140^2 +60^2 - 2(60)(140)cos 110°
= ...
To solve this problem, we can use the concept of vector components. Let's break down the distance traveled into two components: one due west and the other at a bearing of 340 degrees.
The distance traveled due west is 140 km. Since it's traveling directly west, we don't need to calculate any components for this part.
The second distance of 60 km at a bearing of 340 degrees can be resolved into horizontal and vertical components. The horizontal component represents the distance traveled in the east-west direction, and the vertical component represents the distance traveled in the north-south direction.
To find the horizontal component, we need to multiply the distance (60 km) by the cosine of the bearing. The cosine of 340 degrees is the same as the cosine of (360 degrees - 20 degrees), which is approximately 0.9397. So, the horizontal component is 60 km * 0.9397 = 56.38 km (rounded to two decimal places).
To find the vertical component, we need to multiply the distance (60 km) by the sine of the bearing. The sine of 340 degrees is the same as the sine of (360 degrees - 20 degrees), which is approximately -0.3420. Notice the negative sign in the result. So, the vertical component is 60 km * -0.3420 = -20.52 km (rounded to two decimal places).
Now, let's add up the horizontal components and the vertical components separately:
Horizontal component: 140 km (west) + 56.38 km (east) = 196.38 km (rounded to two decimal places).
Vertical component: 0 km (south) + (-20.52 km) = -20.52 km (rounded to two decimal places).
To find the total distance from the airport to the plane, we can use the Pythagorean theorem. The total distance is the square root of the sum of the squares of the horizontal and vertical components:
Total distance = sqrt((196.38 km)^2 + (-20.52 km)^2)
Total distance = sqrt(38480.8544 km^2 + 420.5904 km^2)
Total distance = sqrt(38801.4448 km^2)
Total distance ≈ 197 km (rounded to the nearest whole number).
So, the plane is approximately 197 km away from Vancouver International Airport.