+129899 We can get the answer to 25 by first finding the distance,d, the plane has traveled for 10:24 to 11PM.....since it has been flying for 36 min...we have that
d/36 = 17.91/5
d =36*17.91/ 5 ≈ 128.952 miles
And we can find the (x, y) coordinates of its terminal point as follows
[ 120cos(280) + 128.952cos(69.28) , 120sin(280) + 128.952sin(69.28) ] → x = 66.4612 and y = 2.43453
So....the new trminal point is [ 66.4612, 2.43453]
And the distance [from (0,0) ] is given by ..... sqrt [ 66.4612^2 + 2.43453^2 ] ≈ 66.5058 mi ≈ 67 miles [rounded]
And the bearing is given by [ 90 - arctan(2.43453/66.4612) ] ° = [90 - 2.098]° ≈ 87.902° = 88° [rounded]
So...the new heading (bearing) is 88° at 67 miles
+129899 We can use the Law of Cosines to get tthe answer to 23
We need to find the distance, x, traveled by theplane in 5 minutes....note that the angle betewwn the vectors = 5°
So...we have
x^2 = 120^2 + 105^2 - 2(120)(105)cos(5°)
x = sqrt [ 120^2 + 105^2 - 2(120)(105)cos(5°)] ≈ 17.91 miles
Then...if it traveled 17.91 miles in 5 minutes...it would be traveling at 17.91 * 12 ≈ 214.92mph ≈ 215 mph
To get the answer to 24, let us find the slope between the coordinates end points of the vectors...we have
[ 120cos(280), 120 sin(280)] and ( [ 105cos(285), 105sin(285) ] =
[ 105sin(285) - 120sin(280) ] / [105cos(285) - 120cos(280)] ≈ 2.643
Using the tangent inverse of this to find the angle of the slope, we have arctan(2.643) ≈ 69.28°
And subtracting this from 90 will give us a heading of about 90 - 69.28 = 20.72° ≈ 21°
[I'm still thinking about the answer to 25 ]
+118687 Q24 am just using CPhill's method of putting it all on a grid.
so
\(gradient=\frac{-105sin75--120cos10}{105cos75-120sin10}\\ gradient=\frac{-105sin75+120cos10}{105cos75-120sin10}\\ gradient=2.64\\ tan\theta=2.64\\ \theta\approx 69.28\\ \)
CPhill did this ths fast way LOL
Now 69.28 degrees is the angle between the positive x axis and the direction the plane is flying
but the x axis is in the due east direction whereas direction is taken from the due north direction so
the plane is flying on a bearing of 90-69.28 = 20.72 degrees
So just like CPhill found the plane is flying with a bearing of approximately 21 degrees.
+129899 We can get the answer to 25 by first finding the distance,d, the plane has traveled for 10:24 to 11PM.....since it has been flying for 36 min...we have that
d/36 = 17.91/5
d =36*17.91/ 5 ≈ 128.952 miles
And we can find the (x, y) coordinates of its terminal point as follows
[ 120cos(280) + 128.952cos(69.28) , 120sin(280) + 128.952sin(69.28) ] → x = 66.4612 and y = 2.43453
So....the new trminal point is [ 66.4612, 2.43453]
And the distance [from (0,0) ] is given by ..... sqrt [ 66.4612^2 + 2.43453^2 ] ≈ 66.5058 mi ≈ 67 miles [rounded]
And the bearing is given by [ 90 - arctan(2.43453/66.4612) ] ° = [90 - 2.098]° ≈ 87.902° = 88° [rounded]
So...the new heading (bearing) is 88° at 67 miles
