What is the magnitude of the tension in string 1, if θ1 = 39.85°, θ2 = 64.71° the mass of the sign is 8.7 kg

What is the magnitude of the tension in string 1, if angle 1 = 39.85 degrees, angle 2 = 64.71 degrees, and the mass of the sign is 8.7 kg?

Hello

You draw a parallelogram of forces with the weight force as the vertical diagonal, and the strings, and parallels to the strings, = the sides of the parallelogram. The length of the sides in force units is
the tension in the ropes.
if 39.85° and 64.71° are the angles of the ropes with the vertical, the parallelogram is divided by the diagonal into 2 congruent triangles.
The base side of the triangle = 8.7*9.81 = 85.347 N, the side angles are 39.85° and 64.71, the angle opposed to the base side is 180-39.85-64.71 = 75.44°.
Now find the sides by the sine theorem:
x = tension in string 1 :
x/sin64.71° = 85.347/sin75.44°
x = sin64.71*85.347/sin75.44
x = 79.72 N <--- tension in string 1
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if the angles are with the horizontal, the angles with the vertical are 50.15° and 25.29°.
and the angles opposite to the base side is 104.56°
in this case is x/sin25.29 = 85.347/sin104.56
x = sin25.29*85.347/sin104.56
x = 37.67 N <-- tension in string 1

Regards,

Are the angles from vertical or horizontal? I'll assume from vertical.
(8.7 x 9.8) = 85.26N. Halve it, = 42.63N. normal force for each string.
Tension T1 = 42.63/(cos 39.85) = 55.53N.

Use SINE if angle is from horizontal.