Answer:
[tex]12[/tex].
Step-by-step explanation:
In a right triangle, the tangent of an angle (not the right angle) is the ratio between the length of the side opposite to this angle and that of the side adjacent to this angle. (Opposite over adjacent.)
For example, if [tex]{\rm \triangle EFG}[/tex] is a right triangle where [tex]\angle {\rm F} = 90^{\circ}[/tex]:
[tex]\displaystyle \tan(\angle {\rm E}) = \frac{(\text{side opposite to $\angle {\rm E}$)}}{(\text{side adjacent to $\angle {\rm E}$)}} = \frac{{\rm FG}}{{\rm EF}}[/tex].
Refer to the diagram attached (not to scale.) In this question, the goal is to find [tex]\tan({\rm C})[/tex], which is the ratio between:
- Length of the side opposite to [tex]\angle {\rm C}[/tex], which is also the side opposite to [tex]\angle {\rm A}[/tex], and
- Length of the side adjacent to [tex]{\rm \angle C}[/tex]. which is also the side adjacent to [tex]{\rm \angle B}[/tex].
Let the length of the side adjacent to [tex]\angle {\rm C}[/tex] (and also, adjacent to [tex]\angle {\rm B}\![/tex]) be [tex]1[/tex] unit. Since [tex]\tan({\rm \angle B})[/tex] is the length ratio between the side opposite to [tex]\angle {\rm B}[/tex] and the side adjacent to [tex]\angle {\rm B}\!\![/tex], the length of the side opposite to [tex]\! \angle {\rm B}[/tex] would be [tex](1)\, (\tan({\rm B})) = \tan({\rm B})[/tex] units.
The side opposite to [tex]\angle {\rm B}[/tex] is also the side adjacent to [tex]\angle {\rm A}[/tex]. Similarly, since [tex]\tan({\rm A})[/tex] is the length ratio between the side opposite to [tex]\angle {\rm A}\![/tex] and the side adjacent to [tex]\angle {\rm A}\!\![/tex], the length of the side opposite to [tex]\angle {\rm A}[/tex] would be:
[tex]\begin{aligned} & (\text{length of side opposite to $\angle {\rm A}$})\\ =\; & \tan({\rm A})\, (\text{length of side adjacent to $\angle {\rm A}$}) \\ =\; & \tan({\rm A})\, \tan({\rm B})\end{aligned}[/tex].
Note that the side opposite to [tex]{\angle {\rm A}}[/tex] is also the side opposite to [tex]\angle {\rm C}[/tex]. Hence:
[tex]\begin{aligned} \tan(\angle {\rm C}) &= \frac{(\text{side opposite to $\angle {\rm C}$)}}{(\text{side adjacent to $\angle {\rm C}$)}} \\ &= \frac{\tan({\rm A})\, \tan({\rm B})}{1} \\ &= \tan({\rm A})\, \tan({\rm B})\end{aligned}[/tex].
