Answer: Ground Speed = 91 km/hr, Bearing = 189°
Step-by-step explanation:
Step 1: Draw a picture (see attached) to determine the angle between the given vectors. Notice that I moved the wind vector 180° so the head of the wind vector would line up with the tail of the plane vector. This created an angle of 34° between the plane and wind vectors. Why?
Step 2: Solve for the length of the resultant vector using Law of Cosines
c² = a² + b² - ab cos C
c² = (111)² + (25)² - (111)(25) cos 34°
c² = 12,946 - 4601
c² = 8345
c = 91
Ground speed is 91 km/hr
Step 3: Solve for the bearing of the resultant vector using Law of Sines
[tex]\dfrac{sin\ A}{a}=\dfrac{sin\ C}{c}[/tex]
[tex]\dfrac{sin\ A}{25}=\dfrac{sin\ 34}{91}[/tex]
[tex]sin\ A=\dfrac{25\ sin\ 34}{91}[/tex]
[tex]A=sin^{-1}\bigg(\dfrac{25\ sin\ 34}{91}\bigg)[/tex]
A = 9°
Reminder that we moved the wind vector 180° to create the resultant vector so we need to add 180° to our answer.
Bearing = A + 180°
= 9° + 180°
= 189°
The ground speed and the bearing of the plane are respectively; 109.03 km/h and 166.96º
If we imagine a triangle to depict the question with one side 111 km/h at an angle of 79º to the right of vertical (the plane's speed vector).
At the end, we have a line 25 km/h at an angle of 180 (from north), a vertical down from the end of the plane's vector. This vertex is angle B
This makes a triangle with the interior angle B of 79º.
The 3rd side is the ground speed which we can find using the Cosine rule:
b² = a² + c² - 2•a•c•cos(79)
b² = 111² + 25² - 2•111•25•cos(79)
b² = 11887.01
b = 109.03 km/h
The true course bearing is the ground track. Thus;
Using the Sine rule to find angle A, we have;
(sin(A))/111 = sin(79)/b
Thus,
sin(A) = (111 x sin(79))/109.03
sin(A) = 0.9994
A = sin^(-1)0.9994
A = 87.96º
Ground track = A + 79 = 87.96 + 79 = 166.96º
Read more about Speed and Direction at; https://brainly.com/question/4931057
