Find the sum of vectors A=5m<60degree and B=6m<130degree
How did u come about with cos 110°
The question is complicated
A=5m<60°
B=6m <130°
Solve the question for me
c²=a²+b²-2a cos110°
=25+36-2(5)(6)cos110°
=81.52=c=9.02
Using sine rule we have
SinB/b =Sin110/c
Sin B = b/c X sin110
= 6/9.02 (0.9397) = 0.62446
Therefore. B = 38.64°
c makes an Angle 60° +38.64° =98.6425° with horizontal,that is c = 9.02m<99° .
How come about the 81.52
To find the sum of vectors A and B, we need to break down each vector into its horizontal and vertical components. Then we can add up the corresponding components separately to find the resultant vector.
Let's start with vector A:
Vector A has a magnitude of 5m and an angle of 60 degrees. We can find the horizontal and vertical components using trigonometry.
The horizontal component (A_x) is given by A_x = A * cos(angle), where A is the magnitude and angle is the angle with respect to the x-axis.
A_x = 5m * cos(60 degrees)
A_x = 5m * 0.5
A_x = 2.5m
The vertical component (A_y) is given by A_y = A * sin(angle), where A is the magnitude and angle is the angle with respect to the x-axis.
A_y = 5m * sin(60 degrees)
A_y = 5m * (√3/2) (approximating √3 as 1.732)
A_y ≈ 5m * 1.732
A_y ≈ 8.66m
So the horizontal component of vector A is 2.5m and the vertical component is 8.66m.
Now let's calculate the components of vector B:
Vector B has a magnitude of 6m and an angle of 130 degrees. We will use the same trigonometric formulas.
The horizontal component (B_x) is given by B_x = B * cos(angle).
B_x = 6m * cos(130 degrees)
B_x = 6m * (-0.6428) (approximating cos(130 degrees) as -0.6428)
B_x ≈ -3.857m
The vertical component (B_y) is given by B_y = B * sin(angle).
B_y = 6m * sin(130 degrees)
B_y = 6m * 0.7660 (approximating sin(130 degrees) as 0.7660)
B_y ≈ 4.596m
So the horizontal component of vector B is approximately -3.857m and the vertical component is approximately 4.596m.
Now we can add up the corresponding components:
Horizontal component of the resultant vector (R_x) = A_x + B_x
R_x = 2.5m + (-3.857m) (subtracting since B_x is negative)
R_x = -1.357m
Vertical component of the resultant vector (R_y) = A_y + B_y
R_y = 8.66m + 4.596m
R_y = 13.256m
Finally, we can find the magnitude and angle of the resultant vector using R = √(R_x^2 + R_y^2) and θ = tan^(-1)(R_y / R_x):
Resultant vector (R) = √((-1.357m)^2 + (13.256m)^2)
R ≈ √(1.844m^2 + 175.603m^2)
R ≈ √(177.447m^2)
R ≈ 13.317m
The angle (θ) is given by tan^(-1)(13.256m / -1.357m):
θ ≈ tan^(-1)(-9.747)
θ ≈ -82.61 degrees
Therefore, the sum of vectors A and B is approximately 13.317m at an angle of -82.61 degrees.
x: 5cos60 +6cos130
y: 5sin60 + 6sin130
Pythag..and tan-1