Question

given △def, which is not equal to cos(f)? options: tan(f), sin(d), sin(f), cos(d). triangle def with right angle at e, de=9, ef=9, df=9√2.

Explanation:

Step1: Identify triangle type

△DEF is right - angled at E, with DE = 8, EF = 8, so it's an isosceles right - triangle. Thus, ∠D=∠F = 45°.

Step2: Calculate cos(F)

In right - triangle DEF, \(\cos(F)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{EF}{DF}\). Given \(EF = 8\), \(DF=8\sqrt{2}\), so \(\cos(F)=\frac{8}{8\sqrt{2}}=\frac{1}{\sqrt{2}}\).

Step3: Calculate \(\tan(F)\)

\(\tan(F)=\frac{\text{opposite}}{\text{adjacent}}=\frac{DE}{EF}\). Since \(DE = 8\) and \(EF = 8\), \(\tan(F)=\frac{8}{8}=1=\frac{1}{\sqrt{2}}\times\sqrt{2}\)? Wait, no, \(1=\frac{\sqrt{2}}{\sqrt{2}}\), and \(\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\approx0.707\), \(1
eq\frac{\sqrt{2}}{2}\)? Wait, no, in an isosceles right - triangle with legs 8, hypotenuse \(8\sqrt{2}\). \(\cos(F)=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), \(\tan(F)=\tan(45^{\circ}) = 1\), \(\sin(D)=\sin(45^{\circ})=\frac{\sqrt{2}}{2}\), \(\cos(D)=\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), \(\sin(F)=\sin(45^{\circ})=\frac{\sqrt{2}}{2}\). So \(\tan(F)=1\) is not equal to \(\cos(F)=\frac{\sqrt{2}}{2}\). Wait, let's recalculate:
Wait, \(DE = 8\), \(EF = 8\), right - angled at E. So for angle F: adjacent side is EF = 8, opposite side is DE = 8, hypotenuse is DF=\(\sqrt{8^{2}+8^{2}}=\sqrt{128}=8\sqrt{2}\).
\(\cos(F)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{EF}{DF}=\frac{8}{8\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\tan(F)=\frac{\text{opposite}}{\text{adjacent}}=\frac{DE}{EF}=\frac{8}{8}=1\)
\(\sin(D)\): angle D, opposite side is EF = 8, hypotenuse is \(8\sqrt{2}\), so \(\sin(D)=\frac{EF}{DF}=\frac{8}{8\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\sin(F)\): angle F, opposite side is DE = 8, hypotenuse is \(8\sqrt{2}\), so \(\sin(F)=\frac{DE}{DF}=\frac{8}{8\sqrt{2}}=\frac{\sqrt{2}}{2}\)
\(\cos(D)\): angle D, adjacent side is DE = 8, hypotenuse is \(8\sqrt{2}\), so \(\cos(D)=\frac{DE}{DF}=\frac{8}{8\sqrt{2}}=\frac{\sqrt{2}}{2}\)

So \(\tan(F) = 1\) is not equal to \(\cos(F)=\frac{\sqrt{2}}{2}\), while \(\sin(D)\), \(\sin(F)\), \(\cos(D)\) are equal to \(\cos(F)\).

Answer:

\(\tan(F)\)