A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2
1 answer:
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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,
or
substituting the values in the equation we get,
or
or
X = 26.47°
similarly,
or
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.
