The three forces could be lift drag weight
1st, let's calculate the side AB = c in applying the law of cosine:
AC² = BC²+AC² - 2AB.BC. cos87°)
c²= 10.1²+9.8²-2(10.1)(9.8)(cos(87)
c²=102.01 + 96.04 -196.96cos(87)
c² = 187.74 and c = 13.7
Having calculated the 3rd side, now we can apply Heron's formula for the area:
AREA = √p(p-a)(p-b)(p-c), where p is half the perimeter or p=(a+b+c)/2 =16.8
AREA = √16.8(16.8-10.1)(16.8-9.8)(16.8-13.7)
AREA = 49.42 OR 49.4 unit²
Answer:
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