Question

given right triangle def, what is the value of tan(f)?

Explanation:

Step1: Find side DF using Pythagorean theorem

In right - triangle DEF, by the Pythagorean theorem \(EF^{2}=DE^{2}+DF^{2}\). Given \(DE = 40\) and \(EF=41\), we have \(DF=\sqrt{EF^{2}-DE^{2}}=\sqrt{41^{2}-40^{2}}=\sqrt{(41 + 40)(41 - 40)}=\sqrt{81\times1}=9\).

Step2: Calculate \(\tan(F)\)

The tangent of an angle in a right - triangle is defined as the ratio of the opposite side to the adjacent side. For \(\angle F\), the opposite side to \(\angle F\) is \(DE\) and the adjacent side is \(DF\). So \(\tan(F)=\frac{DE}{DF}\). Since \(DE = 40\) and \(DF = 9\), \(\tan(F)=\frac{40}{9}\).

Answer:

\(\frac{40}{9}\)