Notice that the triangle △ABC is a right triangle. The side AB is the hypotenuse of the triangle, since it is opposed to the right angle C.
The tangent of the angle B is given by the ratio of the lengths of the side opposite to it and the side adjacent to it. Then:
[tex]\tan (B)=\frac{AC}{BC}[/tex]
From the Pythagorean Theorem, we know that:
[tex]AC^2+BC^2=AB^2[/tex]
Substitute the values AC=5 and AB=12 and solve for BC:
[tex]\begin{gathered} 5^2+BC^2=12^2 \\ \Rightarrow BC^2=12^2-5^2 \\ \Rightarrow BC^2=144-25 \\ \Rightarrow BC^2=119 \\ \Rightarrow BC=\sqrt[]{119} \end{gathered}[/tex]
Once we know the value of BC, substitute the lengths of AC and BC to find the tangent of B:
[tex]\tan (B)=\frac{5}{\sqrt[]{119}}[/tex]
Since the square root of 119 is approximately 10.9, then:
[tex]\tan (B)=\frac{5}{10.9}[/tex]
