A 155-N sign is supported by two ropes. One rope pulls up and to the right 26.5° above the horizontal with a tension T1, and the other rope pulls up and to the left 49.5° above the horizontal with a tension T2, as shown in the figure. Find the tensions T1 and T2.

T1*Cos26.5 - T2*Cos49.5 = 0

T1*Cos26.5 = T2*co49.5.
T1 = 0.73T2.

T1*sin26.5 + T2*sin49.5 = 155.
Replace T1 with 0.73T2:
0.73T2*sin26.5 + T2*sin49.5 = 155.
0.324T2 + 0.760T2 = 155.
1.084T2 = 155.
T2 = 143 N.

T1 = 0.73*143 = 104 N.

Well, let's see. We have a sign being supported by two ropes. It's like a high-stakes circus act, but with math instead of trapeze artists.

The first rope is pulling up and to the right at an angle of 26.5° above the horizontal. We'll call the tension in this rope T1. Now, since the sign is not moving up or down, the vertical component of T1 must be equal to the weight of the sign, which is 155 N. Right then, T1 * sin(26.5°) = 155 N.

The second rope is pulling up and to the left at an angle of 49.5° above the horizontal. We'll call the tension in this rope T2. Similarly, the vertical component of T2 must also cancel out the weight of the sign. So, T2 * sin(49.5°) = 155 N.

With these two equations, we can solve for T1 and T2. But let me tell you, my calculations are not my strongest suit. I'm more comfortable juggling jokes than numbers. So, I'm going to need a little bit of assistance here. Could you help me out with the math part?

To find the tensions T1 and T2, we will use vector addition.

Step 1: Resolve the forces into their horizontal and vertical components.
Let T1x be the horizontal component of T1, T1y be the vertical component of T1, T2x be the horizontal component of T2, and T2y be the vertical component of T2.

T1x = T1 * cos(26.5°)
T1y = T1 * sin(26.5°)
T2x = T2 * cos(49.5°)
T2y = T2 * sin(49.5°)

Step 2: Write the equations for the vertical forces and horizontal forces.
In the vertical direction,
T1y + T2y = weight of the sign = 155 N

In the horizontal direction,
T1x - T2x = 0 (since there is no horizontal acceleration)

Step 3: Solve the equations for T1 and T2.
From the first equation:
T1 * sin(26.5°) + T2 * sin(49.5°) = 155

From the second equation:
T1 * cos(26.5°) - T2 * cos(49.5°) = 0

Now we can solve these equations to find T1 and T2.
Divide the first equation by sin(26.5°) and the second equation by cos(26.5°) to isolate T1 and T2:

T1 + (sin(49.5°)/sin(26.5°)) * T2 = (155 / sin(26.5°))
(cos(49.5°)/cos(26.5°)) * T2 - T2 = 0

Simplify the equations further:

T1 + 1.92 * T2 = 321.5
1.19 * T2 - T2 = 0
0.19 * T2 = 321.5
T2 = 321.5 / 0.19
T2 = 1692.11 N

Now substitute the value of T2 into the first equation to find T1:

T1 + 1.92 * 1692.11 = 321.5
T1 = 321.5 - (1.92 * 1692.11)
T1 = 321.5 - 3256.46
T1 = -2934.96 N (negative value indicates the direction)

Therefore, the tensions T1 and T2 are approximately -2934.96 N and 1692.11 N, respectively.

To find the tensions T1 and T2, we can break down the given information into components and use basic trigonometry to solve the problem.

First, let's consider the forces acting on the sign. There are two vertical forces: the weight of the sign (155 N) acting downward and the tension T1 acting upward. There are also two horizontal forces: the tension T2 pulling to the left and the horizontal component of T1 pulling to the right.

Now, let's resolve the forces into vertical and horizontal components:

Vertical components:
- The weight of the sign acts downward and has no vertical component.
- The tension T1 acts upward and has a vertical component equal to T1 * sin(26.5°).
- The tension T2 has no vertical component because it pulls horizontally.

Horizontal components:
- The tension T2 pulls to the left and has a horizontal component equal to T2 * cos(49.5°).
- The tension T1 pulls to the right and has a horizontal component equal to T1 * cos(26.5°).

Since the sign is in equilibrium, the sum of the vertical forces must equal zero and the sum of the horizontal forces must equal zero.

Vertical force balance:
T1 * sin(26.5°) - 155 N = 0

Horizontal force balance:
T2 * cos(49.5°) - T1 * cos(26.5°) = 0

We can now solve the two equations simultaneously to find T1 and T2.

From the vertical force balance equation:
T1 * sin(26.5°) = 155 N
T1 = 155 N / sin(26.5°)
T1 ≈ 366.61 N

Substituting this value of T1 into the horizontal force balance equation:
T2 * cos(49.5°) - (366.61 N) * cos(26.5°) = 0
T2 * cos(49.5°) ≈ (366.61 N) * cos(26.5°)
T2 ≈ ((366.61 N) * cos(26.5°)) / cos(49.5°)
T2 ≈ 277.84 N

Therefore, the tensions T1 and T2 are approximately 366.61 N and 277.84 N, respectively.