in triangle ABC ,AB=2 AC=11, find M<C to the nearest tenth
And what is M supposed to be? Was this question accompanied by a figure? You need to be told an angle or third side length to fully define the triangle.
To find angle M<C in triangle ABC, you can use the Law of Cosines.
The Law of Cosines states that for any triangle with sides a, b, and c, and the opposite angles A, B, C respectively:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we are given AB = 2 and AC = 11. So, a = 2, and b = 11.
Plug the values into the formula:
c^2 = 2^2 + 11^2 - 2(2)(11) * cos(C)
Simplifying:
c^2 = 4 + 121 - 44 * cos(C)
c^2 = 125 - 44 * cos(C)
Now, we need to find the value of cos(C). Rearrange the equation:
cos(C) = (125 - c^2) / 44
To find M<C, we need to find the angle whose cosine is equal to the value we calculated. Calculate cos^(-1)((125 - c^2)/44) to get the value of M<C.
Let's substitute the given lengths into the formula to find cos(C):
cos(C) = (125 - c^2) / 44
cos(C) = (125 - 11^2) / 44
cos(C) = (125 - 121) / 44
cos(C) = 4 / 44
cos(C) ≈ 0.090909091
Now, calculate cos^(-1)(0.090909091) to find M<C:
M<C ≈ cos^(-1)(0.090909091)
M<C ≈ 84.3 degrees (rounded to the nearest tenth)
Therefore, M<C is approximately 84.3 degrees.