Answer :
The angle B in triangle ABC is approximately 5.65 degrees. Angle B in triangle ABC, we can use the Law of Cosines
To find angle B in triangle ABC, we can use the Law of Cosines. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:
[tex]c^2 = a^2 + b^2[/tex] - 2ab * cos(C)
In our case, we have side a = 123.7 cm,
side b = 98.5 cm,
and side c = 154.9 cm.
We want to find angle B, which is opposite side b.
We can rearrange the equation to solve for cos(B):
cos(B) = ([tex]a^2 + b^2 - c^2[/tex]) / (2ab)
Now, we substitute the given values:
cos(B) = ([tex]123.7^2 + 98.5^2 - 154.9^2[/tex]) / (2 * 123.7 * 98.5)
Simplifying the equation gives us:
cos(B) = 0.996056
To find angle B, we can take the inverse cosine ([tex]cos^{-1[/tex]) of 0.996056:
B = [tex]cos^{-1[/tex](0.996056)
Using a calculator, we find:
B ≈ 5.65 degrees
Therefore, angle B in triangle ABC is approximately 5.65 degrees.
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