The components of vector A are Ax and Ay (both positive), and the angle that it makes with respect to the positive x axis is θ. Find the angle θ if the components of the displacement vector A are (a) Ax = 11 m and Ay = 11 m, (b) Ax = 19 m and Ay = 11 m, and (c) Ax = 11 m and Ay = 19 m.

a. tan(theta) = Ay/Ax =11 / 11 = 1,

theta = 45deg.

Ax = 11 + 19 + 11

Ax = 41

Ay = 11 + 11 + 19
Ay = 41

tan(theta) = Ay/Ax
tan(theta) = 41/41
theta = inverse tan x (41/41)
theta = 45 degrees with respect to positive x axis

How did you get theta=45degree

Ax=15m and Ay=15m

Well, well, well, it seems we have ourselves a trigonometry problem here! Don't worry, I'll do my best to clown around with these angles.

(a) For vector A with components Ax = 11 m and Ay = 11 m, we can use good old Pythagoras to find the magnitude of vector A. So, A = √(Ax^2 + Ay^2) = √(11^2 + 11^2) = √(121 + 121) = √(242) = √(22 * 11) = 2√22 m. Now, to find the angle θ, we can use the inverse tangent function: θ = tan^(-1)(Ay/Ax) = tan^(-1)(11/11) = tan^(-1)(1) = 45°.

(b) Moving on to vector B with Ax = 19 m and Ay = 11 m. First, we find the magnitude of vector B using Pythagoras: B = √(19^2 + 11^2) = √(361 + 121) = √(482) = √(2 * 241) ≈ √(2 * 244) = √(4 * 61) = 2√61 m. Then, we find the angle θ: θ = tan^(-1)(Ay/Ax) = tan^(-1)(11/19) ≈ tan^(-1)(0.579) ≈ 30°.

(c) Now, let's tackle vector C with Ax = 11 m and Ay = 19 m. Using Pythagoras, C = √(11^2 + 19^2) = √(121 + 361) = √(482) ≈ √(2 * 244) = √(4 * 61) = 2√61 m. Finally, we find the angle θ: θ = tan^(-1)(Ay/Ax) = tan^(-1)(19/11) ≈ tan^(-1)(1.727) ≈ 59°.

So, to sum it up:
(a) θ ≈ 45°
(b) θ ≈ 30°
(c) θ ≈ 59°

I hope you found my answers amusing! Keep smiling and clowning around with those vectors.

To find the angle θ, we can use the trigonometric functions sine and cosine. Specifically, we can use the following formulas:

θ = arctan(Ay / Ax) (1)

or

θ = arccos(Ax / magnitude of A) (2)

where magnitude of A = sqrt(Ax^2 + Ay^2).

Let's use these formulas to calculate the angles θ for each given case:

(a) For Ax = 11 m and Ay = 11 m,

Step 1: Calculate the magnitude of A:
Magnitude of A = sqrt((11)^2 + (11)^2) = sqrt(242) = 15.56 m

Step 2: Calculate the angle θ using formula (2):
θ = arccos(11 / 15.56)
θ ≈ 42.66 degrees

Therefore, for case (a), the angle θ is approximately 42.66 degrees.

(b) For Ax = 19 m and Ay = 11 m,

Step 1: Calculate the magnitude of A:
Magnitude of A = sqrt((19)^2 + (11)^2) = sqrt(482) = 21.97 m

Step 2: Calculate the angle θ using formula (2):
θ = arccos(19 / 21.97)
θ ≈ 37.95 degrees

Therefore, for case (b), the angle θ is approximately 37.95 degrees.

(c) For Ax = 11 m and Ay = 19 m,

Step 1: Calculate the magnitude of A:
Magnitude of A = sqrt((11)^2 + (19)^2) = sqrt(482) = 21.97 m

Step 2: Calculate the angle θ using formula (2):
θ = arccos(11 / 21.97)
θ ≈ 57.06 degrees

Therefore, for case (c), the angle θ is approximately 57.06 degrees.

By using the given components of vector A and applying the trigonometric formulas, we were able to find the values of θ for each case.