Answer:
m∠A = 39°
Step-by-step explanation:
From the figure attached,
BE ≅ BC
m∠C = 32°
m∠BFD = 103°
In ΔBCE,
m∠E = m∠C = 32° [Since, BE ≅ BC]
m∠E + ∠C + m∠EBC = 180°
32° + 32° + m∠EBC = 180°
m∠EBC = 116°
m∠EBC + m∠EBA = 180° [Linear pair of angles]
116° + m∠EBA = 180°
m∠EBA = 64°
Similarly, m∠AFB + m∠DFB = 180° [Linear pair of angles]
m∠AFB + 103° = 180°
m∠AFB = 77°
Now in ΔAFB,
m∠A + m∠AFB + m∠FBA = 180° [Sum of internal angles of a triangle = 180°]
m∠A + 77° + 64° = 180°
m∠A = 39°
Add -46 and -46 to the final answer
Answer:
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Step-by-step explanation:
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The sum of the k for k=1 to k="n"=(n*(n+1))/2 , where "n" is the total integer.
In this problem n = 200
Sum = (200*201)/2 = 20,100
You need to find, first, the horizontal component of the inclined force (call it Wx), using an angle of 15 degrees (which is the angle between the inclined plane and the horizontal).
=> cos(15°) = Wx / 2.5 lb => Wx = 2.5 lb * cos(15°).
Second, you can find the inclined force , W, needed to produce the horizontal component Wx.
cos(33°) = Wx / W => W = Wx / cos (33°) = 2.5 lb*cos(15°) / cos(33°) = 2.88 lb
Answer: 2.88 lb