In Triangle ABC, a=7, b=6, and c=8. find angle B to the nearest degree?
To find angle B in triangle ABC, you can use the Law of Cosines:
c^2 = a^2 + b^2 - 2ab*cos(B)
Substituting the given values, we get:
8^2 = 7^2 + 6^2 - 2(7)(6)*cos(B)
64 = 49 + 36 - 84*cos(B)
64 = 85 - 84*cos(B)
Simplifying further:
84*cos(B) = 85 - 64
84*cos(B) = 21
cos(B) = 21/84
cos(B) ≈ 0.25
To find the angle, take the inverse cosine (cos^-1) of 0.25.
B ≈ cos^-1(0.25)
B ≈ 75.52 degrees
Therefore, angle B is approximately 76 degrees (to the nearest degree).
To find angle B in Triangle ABC, we can use the Law of Cosines, which states that for any triangle with side lengths a, b, and c and opposite angles A, B, and C respectively:
c^2 = a^2 + b^2 - 2ab*cos(C)
In this case, we have a=7, b=6, and c=8. We want to find angle B. Let's substitute these values into the formula:
8^2 = 7^2 + 6^2 - 2*7*6*cos(B)
64 = 49 + 36 - 84*cos(B)
64 = 85 - 84*cos(B)
Now, rearrange the equation to isolate the cosine of angle B:
84*cos(B) = 85 - 64
84*cos(B) = 21
cos(B) = 21/84
cos(B) = 1/4
To find angle B, we need to take the inverse cosine (cos^-1) of 1/4. Let's calculate it:
B = cos^-1(1/4)
Using a calculator or a trigonometric table, we find that cos^-1(1/4) is approximately 75.52 degrees.
Therefore, to the nearest degree, angle B in Triangle ABC is 76 degrees.