What is the area of ∆ABC to the nearest tenth of a square meter?

The remaining side using cosine theorem is:
BC ^ 2 = 36 ^ 2 + 36 ^ 2 - 2 * 36 * 36 * cos (65)
BC = root (1496.573466)
BC = 38.68557181
Then, for the area we have:
A = root (s (s-a) * (s-b) * (s-c))
s = (a + b + c) / 2
Substituting:
s = (36 + 36 + 38.68557181) / 2
s = 55.34278591
A = root (55.34278591 * (55.34278591-36) * (55.34278591-36) * (55.34278591-38.68557181))
A = 587.2874463
Rounding off we have:
A = 587.3
Answer:
A = 587.3

aencabo aencabo · Answer 2 · 2018-05-13T14:06:48+02:00

aencabo aencabo

13-05-2018

The answer would be: nswer: A = 587.3

Solution:
BC ^ 2 = 36 ^ 2 + 36 ^ 2 - 2 * 36 * 36 * cos (65) BC = root (1496.573466) BC = 38.68557181 Then, for the area we have: A = root (s (s-a) * (s-b) * (s-c)) s = (a + b + c) / 2 Substitute: s = (36 + 36 + 38.68557181) / 2 s = 55.34278591 A = root (55.34278591 * (55.34278591-36) * (55.34278591-36) * (55.34278591-38.68557181)) A = 587.2874463 A = 587.3

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