Farhad George and Tom each did a test here are the marks for the test George drew this bar chart to show the mask they got the b
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Third leg.
The crew flies at a speed of 560 mi/h in direction N-20°-E.
The wind has a speed of 35 mi/h and a direction S-10°-E.
We then can draw this as:
We have to add the two vectors to find the actual speed and direction.
We will start by adding the x-coordinate (W-E axis):
and the y-coordinate (S-N axis) is:
Then, the actual speed vector is v3=(197.61, 491.76).
The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].
Then, we have to calculate the displacement in 20 minutes using the actual speed vector.
We can calculate the movement in each of the axis. For the x-axis:
NOTE: 20 minutes represents 1/3 of an hour.
We can do the same with the y-coordinate:
The final position is R3 = (282.53, 331.59).
To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:
The distance can be calculated as if it was a right triangle:
The angle, from E to N, can be calculated as:
If we want to express it from N to E, we substract the angle from 90°:
Answer: the final location can be represented with the vector (282.53, 331.59).
1) The distance from the origin is 435.63 miles and
2) the direction is N-40°-E.
Answer:
The answer is;
54.4
Step-by-step explanation:
The given parameter are;
Triangle ΔFGH = Right triangle
The length of segment = 48
The measure of ∠HFG = 28°
The measurement required = The measure of segment = x
The measure of ∠FHG = 90° angle opposite hypotenuse side of a right triangle
Therefore, ∠FGH = 180° - ∠HFG - ∠FHG = 180° - 90° - 28° = 62°
∠FGH = 62°
By sine rule, we have;
/(sin ∠FGH) =
/(sin(∠FHG)
By substituting the known values, we have;
48/(sin 62°) = x/(sin(90°)
sin(90°) = 1, therefore, we have;
x/1 = x = 48/(sin 62°) = 54.4 (by rounding the answer to the nearest tenth)
x = 54.4