Answer:
The last two bearings are
49.50° and 104.02°
Explanation:
Applying the Law of cosine (refer to the figure attached):
we have
x² = y² + z² - 2yz × cosX
here,
x, y and z represents the lengths of sides opposite to the angels X,Y and Z.
Thus we have,
![cos X=\frac{x^2-y^2-z^2}{-2yz}](https://tex.z-dn.net/?f=cos%20X%3D%5Cfrac%7Bx%5E2-y%5E2-z%5E2%7D%7B-2yz%7D)
or
![cos X=\frac{y^2 + z^2-x^2}{2yz}](https://tex.z-dn.net/?f=cos%20X%3D%5Cfrac%7By%5E2%20%2B%20z%5E2-x%5E2%7D%7B2yz%7D)
substituting the values in the equation we get,
![cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}](https://tex.z-dn.net/?f=cos%20X%3D%5Cfrac%7B2900%5E2%20%2B%203700%5E2-1700%5E2%7D%7B2%5Ctimes%202900%5Ctimes%203700%7D)
or
![cos X=0.8951](https://tex.z-dn.net/?f=cos%20X%3D0.8951)
or
X = 26.47°
similarly,
![cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}](https://tex.z-dn.net/?f=cos%20Y%3D%5Cfrac%7B1700%5E2%20%2B%203700%5E2-2900%5E2%7D%7B2%5Ctimes%201700%5Ctimes%203700%7D)
or
![cos Y=0.649](https://tex.z-dn.net/?f=cos%20Y%3D0.649)
or
Y = 49.50°
Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°
The bearing of 2 last legs of race are angels Y and Z.
Answer:
R = 715.4 N
L = 166.6 N
Explanation:
ASSUME the painter is standing right of center
Let L be the left rope tension
Let R be the right rope tension
Sum moments about the left end to zero. Assume CCW moment is positive
R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0
R = 715.4 N
Sum moments about the right end to zero
20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0
L = 166.6 N
We can verify by summing vertical forces
116.6 + 715.4 - (70 + 20)(9.8) ?=? 0
0 = 0 checks
If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.
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