Asked by Anonymous
A triangular parcel of land has borders of lengths 55 meters, 85 meters, and 100 meters. Find the area of the parcel of land.
Answers
Answered by
Steve
pick any angle (say, A) and find it via
a^2 = b^2 + c^2 - 2bc cosA
Now the area is given by 1/2 bc sinA
a^2 = b^2 + c^2 - 2bc cosA
Now the area is given by 1/2 bc sinA
Answered by
Reiny
checking for right-angled triangle:
is 100^2 = 55^2 + 85^2 ?
NO
so let's find the smallest angle by the cosine law:
55^2 = 100^2 + 85^2 - 2(100)(85)cosØ
cosØ = 14200/17000 = .83529..
Ø = 33.35..° , (I stored it for accuracy on my calculator
So area = (1/2)(85)(100)sin 33.35..° = appr 2336.66
or using Heron's formula
s = (1/2)(100+85+55) = 120
s-a = 120-100 = 20
s-b = 120 - 85 = 35
s-c = 120-55 = 65
area = √(120*20*35*65) = appr 2336.66
is 100^2 = 55^2 + 85^2 ?
NO
so let's find the smallest angle by the cosine law:
55^2 = 100^2 + 85^2 - 2(100)(85)cosØ
cosØ = 14200/17000 = .83529..
Ø = 33.35..° , (I stored it for accuracy on my calculator
So area = (1/2)(85)(100)sin 33.35..° = appr 2336.66
or using Heron's formula
s = (1/2)(100+85+55) = 120
s-a = 120-100 = 20
s-b = 120 - 85 = 35
s-c = 120-55 = 65
area = √(120*20*35*65) = appr 2336.66
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