An aircraft takes off at an angle of 5.2 degrees to the ground.How high is it when it has moved 2000m horizontally from its take-off point
plane: h/2000 = tan 5.2°
for the other, fix the typos, draw a diagram and use the law of cosines/sines.
A town B is due north of A.A third town C is 10km on a bearing 020 degrees from A.If B is on a bearing of 290 degrees from B.calculate (a) BC (b) AB
A town B is due north of A.A third town C is 10km on a
bearing 020 degrees from A.If B is on a bearing of 290
degrees from B.calculate (a) BC (b) AB
Please how will it be solved
How can i solv it
Pls solve this
To find the height of the aircraft when it has moved 2000m horizontally from its take-off point, we can use trigonometry and the concept of right triangles.
First, let's define the given information:
- The angle of elevation (angle between the line of sight and the horizontal ground) is 5.2 degrees.
- The horizontal distance traveled is 2000m.
We can use the tangent function to calculate the height of the aircraft. The tangent of an angle is defined as the ratio of the opposite side to the adjacent side in a right triangle.
In this case, the opposite side represents the height of the aircraft, and the adjacent side represents the horizontal distance traveled. Therefore, we can write the equation:
tan(angle) = height / distance
Substituting the known values:
tan(5.2 degrees) = height / 2000m
To solve for the height, we need to isolate it. Rearranging the equation, we have:
height = tan(5.2 degrees) * 2000m
Now, let's calculate the height using this equation:
height = tan(5.2 degrees) * 2000m
Using a scientific calculator, determine the tangent of 5.2 degrees (make sure your calculator is set to degrees mode). Let's assume the tangent is approximately 0.0908. Plugging this value into the equation:
height = 0.0908 * 2000m
height ≈ 181.6m
Therefore, when the aircraft has moved 2000m horizontally from its take-off point, it is approximately 181.6 meters high.