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Firdavs [7]
3 years ago
9

Secants ⎯⎯⎯⎯⎯⎯⎯⎯⎯ and ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ intersect in the exterior of a circle. DJH measures 191° and EG measures 53°. Determine the mea

sure of ∠ EFG.

Mathematics
1 answer:
Alik [6]3 years ago
4 0

Answer:

m\angle EFG =69\degree

Step-by-step explanation:

m\angle EFG =\frac{1}{2}(191- 53)\degree

m\angle EFG =\frac{1}{2}\times 138\degree

m\angle EFG =69\degree

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SOLVED
Nastasia [14]
5)
a. The equation that describes the forces which act in the x-direction: 
<span>     Fx = 200 * cos 30 </span>
<span>
b. The equation which describes the forces which act in the y-direction: </span>
<span>     Fy = 200 * sin 30 </span>

<span>c. The x and y components of the force of tension: </span>
<span>    Tx = Fx = 200 * cos 30  </span>
<span>    Ty = Fy = 200 * sin 30 </span>

d.<span>Since desk does not budge, </span><span>frictional force = Fx
                                                                        = 200 * cos 30 </span>

<span>                                                 Normal force </span><span>= 50 * g - Fy
                                                                       = 50 g - 200 * sin 30 
</span>____________________________________________________________
6)<span> Let F_net = 0</span>
a. The equation that describes the forces which act in the x-direction: 
    (200N)cos(30) - F_s = 0

b. The equation that describes the forces which act in the y-direction:
    F_N - (200N)sin(30) - mg = 0

c. The values of friction and normal forces will be:
     Friction force= (200N)cos(30),
     
The Normal force is not 490N in either case...
Case 1 (pulling up)
F_N = mg - (200N)sin(30) = 50g - 100N = 390N

Case 2 (pushing down)
F_N = mg + (200N)sin(30) = 50g + 100N = 590N
4 0
3 years ago
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