00:01
In this question, it is given that a plane is flying at a speed of 320 miles per hour on a bearing of an 70 degree e.
00:11
Its ground speed is 370 miles per hour and its true course is 30 degree.
00:18
We need to find the speed to the nearest mile per hour and the direction angle to the nearest tenth of a degree of the wind.
00:29
So let's see how to solve this question.
00:30
So, based on the available information, we can draw the vector diagram for the plane speed and the wind speed as shown below.
00:42
So this is a vector diagram.
00:44
Here, v represents the plane speed, w represents the wind speed and v plus w represents the resultant speed.
00:56
Now for the vector v, the direction angle theta from the horizontal can be calculated.
01:04
As theta is equal to 90 degree minus 70 degree.
01:13
Hence this will be equal to 20 degree.
01:18
We know that if we have the magnitude and direction angle then a non -zero vector can be represented as v is equal to magnitude v.
01:30
Cosine theta i plus magnitude v sine theta j.
01:40
Therefore, the velocity of the plane v can be written as 320 miles per hour, which is given cosine 20 degree i plus 320 sine 20 degree j.
02:04
And when we further calculate this, we get velocity of the plane vector v is equals to 300 .70i plus 109 .45j.
02:26
Since it is given that the ground speed is 370 miles per hour and its true course is 30 degree.
02:35
Therefore, the vector representation of v plus w can be written as 370 miles per hour, which is given, cosine, this angle is 30 degree, 30 degree i plus 30070 sign 30 degree j.
03:06
When you further calculate this, we get v plus w is equals to 320 .43i plus 185j.
03:25
This was equation 1 and this one is equation 2.
03:37
And now let's write the vector representation of velocity of wind...