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In triangle ABC, a-64, b-18, and c-63. Find the value of B.​

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MrRoyalidn

Triangles are shapes with three sides and angles

The value of B is 16.3 degrees

How to determine the value of B

The given parameters are:

a = 64

b = 18

c = 63

Next, we apply the following cosine formula

[tex]b^2 = a^2 + c^2 - 2ac\ cos(B)[/tex]

So, we have:

[tex]18^2 = 64^2 + 63^2 - 2 * 64 * 63 \ cos(B)[/tex]

Evaluate the exponents

[tex]324 = 8065 - 8064\ cos(B)[/tex]

Collect like terms

[tex]324 - 8065 = - 8064\ cos(B)[/tex]

[tex]-7741 = - 8064\ cos(B)[/tex]

Solve for cos (B)

[tex]\cos(B) =0.9599[/tex]

Take the arccos of both sides

[tex]B =cos^{-1}(0.9599)[/tex][tex]B = 16.3[/tex]

Hence, the value of B is 16.3 degrees

Read more about triangles at:

https://brainly.com/question/17972372

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