An aeroplane flies due north from ikeja airport for 500km, it then flies on a bearing of 060 degree for a further distance of 300km before overflying a road junction. Calculate: the bearing of the aeroplane from ikeja airport at this instant
North 500 + 300 cos 60 = 650
East 300 sin 60 = 260
tan bearing from origin = East/North =260/650 = 0.4
bearing = 21.8 deg
By the way you fly on a heading, not a bearing. You take a bearing on something like a lighthouse or a plane. Math teachers do not fly airplanes or sail ships :)
To find the bearing of the airplane from Ikeja airport at the moment it overflies the road junction, we can use trigonometry and the concept of bearings.
First, let's visualize the scenario. The airplane starts at Ikeja airport and flies due north for 500km. Then, it changes its bearing by 060 degrees and continues flying for another 300km until it reaches the road junction.
Now, we need to find the angle that represents the bearing of the airplane from Ikeja airport at this point.
Using trigonometry, we can form a right triangle with the initial flight (due north) as the vertical side, the change in bearing (060 degrees) as the horizontal side, and the hypotenuse as the distance traveled to the road junction (300km).
Since we have a right triangle, we can use the tangent function to find the angle:
tan(θ) = opposite/adjacent = 300/500
To solve for θ, we take the inverse tangent of both sides:
θ = tan^(-1)(300/500)
Using a calculator, we find that θ is approximately 30.96 degrees.
Therefore, the bearing of the airplane from Ikeja airport at the moment it overflies the road junction is 30.96 degrees.
To determine the bearing of the airplane from Ikeja Airport at the instant it overflies the road junction, we can use trigonometry and the given information.
First, let's visualize the situation:
- The airplane initially flies due north for 500km from Ikeja Airport.
- It then changes its direction and flies on a bearing of 060 degrees for a further distance of 300km before reaching the road junction.
Now, we need to find the angle between the airplane's initial path and the line connecting Ikeja Airport to the road junction.
We can create a triangle with sides representing the airplane's flight paths:
- Side A represents the 500km northward flight from Ikeja Airport.
- Side B represents the 300km flight on a bearing of 060 degrees.
Using the Law of Cosines, we can find the third side, C, in the triangle:
C^2 = A^2 + B^2 - (2 * A * B * cos(C))
Since we need to find the bearing of the airplane from Ikeja Airport, we are interested in the angle C formed by sides A and B.
Now, let's calculate the value of C using the Law of Cosines:
C^2 = 500^2 + 300^2 - (2 * 500 * 300 * cos(C))
Next, we can solve for C:
C^2 = 250000 + 90000 - (300000 * cos(C))
C^2 = 340000 - (300000 * cos(C))
At this point, we need to use algebra or numerical methods to find the value of C. Once we have the value of C, we can determine the bearing of the airplane from Ikeja Airport using the following steps:
1. Subtract C from 180 degrees to find the interior angle in the triangle opposite side A (the northward flight). Let's call this angle D.
D = 180 - C
2. Add 60 degrees to the angle D to find the bearing of the airplane from Ikeja Airport at the instant it overflies the road junction.
If you have the value of C, you can plug it into the formula above and follow these steps to calculate the bearing of the airplane from Ikeja Airport.
the junction is
500+300cos60° = 650 km north
and 300 sin60° = 150√3 km east
so the bearing of the plane from the airport is θ, where
tanθ = 150√3/650
now use your calculator to find θ