7 m/s on a bearing of 30° and 2 m/s from a bearing of 343°. How would you find the direction and magnitude using both Sine and Cosine Law? I know how to answer this question using Components....
NOT a BEARING, they mean a HEADING
(mathematicians drive me nuts).
A bearing is the direction you look, not the drirection you head.
Anyway:
Of couse do it with components but
Call angle at origin A, 0 deg is north
it is 30 + (360-343) = 30 + 17 = 47 deg
call Angle in NW quadrant B (quad3) and Angle in NE quadrant(quad1) = C
now
BC^2 = AB^2 + AC^2 -2 AB AC cos 47
but
AB = 2 and AC = 7
so
BC^2 = 4 + 49 - 2 (2)(7)cos 47
solve for BC, now have all 3 sides
then use law of sines to get angles
like
sin C /2 = sin A/BC and Sin B / 7 = same thing
then you have all the sides and all the angles.
To find the direction and magnitude of the resultant velocity using both the Sine and Cosine Laws, you need to break down the given velocities into their x-component and y-component vectors. Then, you can apply the Sine and Cosine Laws to find the direction and magnitude.
Let's begin by calculating the x-component and y-component of each given velocity:
1. Velocity of 7 m/s on a bearing of 30°:
x-component: 7 * cos(30°) = 7 * √3/2 ≈ 6.06 m/s (in the positive x-axis direction)
y-component: 7 * sin(30°) = 7 * 1/2 = 3.5 m/s (in the positive y-axis direction)
2. Velocity of 2 m/s from a bearing of 343°:
x-component: 2 * cos(343°) (convert to positive angle by adding 360°) = 2 * cos(23°) ≈ 1.77 m/s (in the positive x-axis direction)
y-component: 2 * sin(343°) (convert to positive angle by adding 360°) = 2 * sin(23°) ≈ 0.77 m/s (in the positive y-axis direction)
Now, let's find the resultant x-component and y-component by summing up the corresponding components:
Resultant x-component: 6.06 m/s + 1.77 m/s = 7.83 m/s (in the positive x-axis direction)
Resultant y-component: 3.5 m/s + 0.77 m/s = 4.27 m/s (in the positive y-axis direction)
Next, we can use the Sine and Cosine Laws to find the magnitude and direction of the resultant velocity:
Magnitude (R): R = √(Rx^2 + Ry^2)
Direction (θ): θ = arctan(Ry/Rx) = arctan(4.27/7.83)
Plugging in the values:
Magnitude (R) ≈ √(7.83^2 + 4.27^2) ≈ √(61.17 + 18.23) ≈ √79.4 ≈ 8.91 m/s (rounding to two decimal places)
Direction (θ) ≈ arctan(4.27/7.83) ≈ 29.3° (rounding to one decimal place)
Therefore, the direction and magnitude of the resultant velocity using both the Sine and Cosine Laws are approximately 8.91 m/s at a bearing of 29.3°.