Piece of a wire

A piece of wire is bent into the shape of a triangle. Two sides have lengths of 24 inches and 21 inches.
The angle between these two sides is 55°. What is the length of the third side to the nearest hundredth of an inch?


A: The length of the third side is approximately ____ inches.

Final Answer:

c =  20.9484 in

Step-by-step explanation:

$$\begin{gathered} a=24\text{in} \\ b=21\text{in} \\ \gamma =55{}^{\circ } \\ {c}^2=a^2+b^2-2ab\cos \gamma \\ c=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos \gamma }=\sqrt{a^2+b^2-2\cdot a\cdot b\cdot \cos 55°}=\sqrt{24^2+21^2-2\cdot 24\cdot 21\cdot \cos 55°}=\sqrt{24^2+21^2-2\cdot 24\cdot 21\cdot 0.573576}=20.948=20.9484\text{in} \end{gathered}$$

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