To the nearest tenth of an inch, what is the length of side AB in triangle ABC with side BC measuring 15 inches, side AC measuring 20 inches, and angle ACB measuring 112 degrees?

To find the length of side AB in triangle ABC given the lengths of sides BC and AC as well as the measure of angle ACB, we can use the Law of Cosines. The Law of Cosines states that in any triangle, ( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ), where ( c ) is the side opposite angle ( C ), and ( a ) and ( b ) are the lengths of the other two sides.

In this scenario, we can set:

• ( a = BC = 15 ) inches

• ( b = AC = 20 ) inches

• ( C = ACB = 112^\circ )

We want to find the length of side AB, which can be denoted as ( c ). Applying the Law of Cosines:

[

c^2 = a^2 + b^2 - 2ab \cdot \cos(C)

]

Substituting the values we have:

[

c^2 = 15^2 + 20^2 - 2 \cdot 15 \cdot 20 \cdot \cos(112^\circ)