(40 points) I need some help please. An answer definition is appreciated as well.
2 answers:
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If you're going to use technology to get help, you may as well use technology that can give you the answer immediately. A triangle solver (or your graphing calculator) can tell you
∠A ≈ 29.5°
∠B ≈ 115.4°
∠C ≈ 35.1°
You can rearrange the formula of the Law of Cosines to give you the angles unambiguously. (It often works best to start with the largest angle.) That law tells you
b² = a² + c²- 2ac·cos(B)
B = arccos((a² + c² - b²)/(2ac))
B = arccos((6² + 7² - 11²)/(2·6·7))
B = arccos(-36/84) ≈ 115.377°
Similarly,
A = arccos((b²+c²-a²)/(2bc)) = arccos(134/154) ≈ 29.526°
C = arccos((a²+b²-c²)/(2ab)) = arccos(108/132) ≈ 35.097°
Of course, once you have 2 angles you can find the third by subtracting their sum from 180°.
Once you have one angle, you can also use the Law of Sines to find another.
a/sin(A) = b/sin(B)
A = arcsin(a/b·sin(B)) = arcsin(6/11·sin(115.377°))
≈ arcsin(.492822) ≈ 29.526°
_____
Always maintain intermediate results to the full precision your calculator offers. Only round the final answer.
∠A ≈ 29.5°
∠B ≈ 115.4°
∠C ≈ 35.1°
You can rearrange the formula of the Law of Cosines to give you the angles unambiguously. (It often works best to start with the largest angle.) That law tells you
b² = a² + c²- 2ac·cos(B)
B = arccos((a² + c² - b²)/(2ac))
B = arccos((6² + 7² - 11²)/(2·6·7))
B = arccos(-36/84) ≈ 115.377°
Similarly,
A = arccos((b²+c²-a²)/(2bc)) = arccos(134/154) ≈ 29.526°
C = arccos((a²+b²-c²)/(2ab)) = arccos(108/132) ≈ 35.097°
Of course, once you have 2 angles you can find the third by subtracting their sum from 180°.
Once you have one angle, you can also use the Law of Sines to find another.
a/sin(A) = b/sin(B)
A = arcsin(a/b·sin(B)) = arcsin(6/11·sin(115.377°))
≈ arcsin(.492822) ≈ 29.526°
_____
Always maintain intermediate results to the full precision your calculator offers. Only round the final answer.