Question

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

Answer 1

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

$b^2 = a^2 + c^2 - 2ac\ cos(B)$

So, we have:

$18^2 = 64^2 + 63^2 - 2 * 64 * 63 \ cos(B)$

Evaluate the exponents

$324 = 8065 - 8064\ cos(B)$

Collect like terms

$324 - 8065 = - 8064\ cos(B)$

$-7741 = - 8064\ cos(B)$

Solve for cos (B)

$\cos(B) =0.9599$

Take the arccos of both sides

$B =cos^{-1}(0.9599)$$B = 16.3$

Hence, the value of B is 16.3 degrees

Read more about triangles at:

brainly.com/question/17972372