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Triangle ABC has the following side lengths: \(b = 98.5 \, \text{cm}\), \(a = 123.7 \, \text{cm}\), and \(c = 154.9 \, \text{cm}\).

Find angle \(B\).

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.

To know more about inverse cosine visit :

https://brainly.com/question/20306331

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