Answer:
[tex]\tan F = \dfrac{\sqrt{390}}{15}[/tex]
Step-by-step explanation:
To find the tangent of ∠F in right triangle EFG, we can use the tangent trigonometric ratio:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=\dfrac{O}{A}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$O$ is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\end{array}}[/tex]
The tangent trigonometric ratio is defined as the ratio of the length of the side opposite an acute angle in a right triangle to the length of the side adjacent to that angle.
In this case:
- The angle is F.
- The side opposite angle F is side EG.
- The side adjacent angle F is side EF.
Therefore:
- [tex]\theta = F[/tex]
- [tex]O = EG = \sqrt{26}[/tex]
- [tex]A = EF = \sqrt{15}[/tex]
Substitute the values into the tangent ratio:
[tex]\tan F = \dfrac{\sqrt{26}}{\sqrt{15}}[/tex]
To write this in its simplest form, we can rationalize the denominator by multiplying both the numerator and denominator by √15. This eliminates the square root in the denominator:
[tex]\tan F = \dfrac{\sqrt{26}\times\sqrt{15}}{\sqrt{15}\times \sqrt{15}} \\\\\\ \tan F = \dfrac{\sqrt{26\times15}}{15} \\\\\\ \tan F = \dfrac{\sqrt{390}}{15}[/tex]
Therefore, the tangent of angle F in simplified, rationalized form is:
[tex]\Large\boxed{\boxed{ \tan F = \dfrac{\sqrt{390}}{15} }}[/tex]
