What is the area of a triangle that has a heaight of 5 feet and a base length of 4 feet
2 answers:
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We'll use the Law of Cosines
We'll call
side a = length 5
side b = length 8
side c = length 10
and their corresponding angles as A, B and C
"C" would be the largest angle
cos (C) = (a^2 + b^2 - c^2) / 2 * a * b
cos (C) = (5^2 + 8^2 - 10^2) / 2 * 5 * 8
cos (C) = 25 + 64 -100 / 80
cos (C) = -11 / 80
cos (C) = <span> <span> <span> -0.1375 </span> </span> </span>
arc cosine ( <span> <span> <span> -0.1375 </span> </span> </span> ) = 97.903 Degrees
The largest angle is 97.903 Degrees
We'll call
side a = length 5
side b = length 8
side c = length 10
and their corresponding angles as A, B and C
"C" would be the largest angle
cos (C) = (a^2 + b^2 - c^2) / 2 * a * b
cos (C) = (5^2 + 8^2 - 10^2) / 2 * 5 * 8
cos (C) = 25 + 64 -100 / 80
cos (C) = -11 / 80
cos (C) = <span> <span> <span> -0.1375 </span> </span> </span>
arc cosine ( <span> <span> <span> -0.1375 </span> </span> </span> ) = 97.903 Degrees
The largest angle is 97.903 Degrees
