given right triangle def, what is the value of tan(f)?\n$\frac{9}{41}$\n$\frac{40}{41}$\n$\frac{40}{9}$\n$\fr…
given right triangle def, what is the value of tan(f)?\n$\frac{9}{41}$\n$\frac{40}{41}$\n$\frac{40}{9}$\n$\frac{41}{9}$
Answer
Explanation:
Step1: Recall tangent definition
For an acute angle in a right triangle, $\tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}$
Step2: Identify sides for $\angle F$
Opposite side to $\angle F$: $DE = 40$; Adjacent side to $\angle F$: $DF = 9$
Step3: Calculate $\tan(F)$
$\tan(F) = \frac{\text{opposite}}{\text{adjacent}} = \frac{40}{9}$
Answer:
$\boldsymbol{\frac{40}{9}}$ (corresponding to the third option: $\boldsymbol{\frac{40}{9}}$)