Question

Anybody who knows Pre Calc?? My last offer here Solution PLS!

Answer 1

Answers:

• true speed = 26.935546 knots
• true bearing = 102.265139 degrees

The values are approximate. Round them however you need to.

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Explanation:

The diagram you have drawn is correct if you're only considering one current. Specifically current #1 which has the bearing of 290 degrees moving at 25 knots.

The red angle is 290 degrees. The remaining portion that is needed to get a full 360 degrees is 360-290 = 70. This adds to the blue portion in quadrant 1, which is 90, and we get 70+90 = 160. So the blue angle is 160 degrees.

The vector for current #1 is <25*cos(160), 25*sin(160)>

This approximates to <-23.4923155, 8.5505036>

All of this is for current #1.

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Vector #2 is much more simple in terms that there aren't messy decimal approximations here. The direction is due north, which means that the vector is pointing straight up toward 90 degrees.

We can then say: <12*cos(90), 12*sin(90)> = <12*0, 12*1> = <0, 12>

Vector #2 is <0,12>

When you wrote <12,0>, you were on the right track, but you had the coordinates swapped in the wrong order.

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Then we have to account for a wind going 4 knots in the southwest direction. This is at the angle 225 degrees (we rotate 180 degrees plus another 45 to get 225). So,

vector #3 = <4*cos(225), 4*sin(225)> = <-2.8284271, -2.8284271>

This vector's coordinates are approximate

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To recap everything so far, vectors 1 through 3 are the following

• p = <-23.4923155, 8.5505036>
• q = <0, 12>
• r = <-2.8284271, -2.8284271>

I'm using variables to represent the vectors at this point, so we can then add them up to get...

s = p+q+r

s = <-23.4923155, 8.5505036> + <0,12> + <-2.8284271, -2.8284271>

s = <-23.4923155+0+(-2.8284271), 8.5505036+0+(-2.8284271)>

s = <-26.3207426, 5.7220765>

Vector s represents the sum of the three vectors p, q, and r. The two currents and the wind ultimately combine together to push the boat in the direction and along the length of vector s.

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Now that we know the coordinates of vector s, we use them to find the true speed and true bearing of the boat.

speed = sqrt(a^2 + b^2)

speed = sqrt( (-26.3207426)^2 + (5.7220765)^2 )

speed = 26.935546 knots, which is approximate

angle = arctan(b/a)

angle = arctan(5.7220765/(-26.3207426))

angle = -12.265139

Add 360 to this angle to find that we get to the coterminal angle of -12.265139+360 = 347.734861 degrees

This angle is in quadrant 4, which is the southeast quadrant. If we rotate another 360-347.734861 = 12.265139 degrees, going counterclockwise, then we'll face directly east. Then another 90 degree counterclockwise rotation has us face directly north.

In total, the true bearing is approximately 90+12.265139 = 102.265139 degrees

So we'll start aiming north, and then rotate roughly 102.265139 degrees toward the east to get to the true bearing.