A line segment has end points R(-2, - 2) and S(4, 1). What are the coordinates of the point that is 2/3 of the way from R to S o
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9² = 12² + 15² - 2 (12) (15) cos (B)
81 = 144 + 225 - 360 cos(B)
81 = 369 - 360 cos (B)
360 cos (B) = 369 - 81
360 cos (B) = 288
cos (B) = 0.8
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Answer: Cosine B = 0.8
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12² = 15² + 9² - 2 (15)(9) cos (A)
144 = 225 + 81 - 270 Cos A
144 = 306 - 270 Cos A
270 Cos A = 162
Cos A = 3/5 or 0.6
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Answer: Cosine Angle A = 3/5
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.
