Chapter 2 - Acute Angles and Right Triangles - Section 2.3 Finding Trigonometric Function Values Using a Calculator - 2.3 Exercises - Page 63: 14

$\cot$(-512$^{\circ}$ 20') $\approx$ 1.90714493

$\cot$(-512$^{\circ}$ 20') First, convert from minutes to degrees -512$^{\circ}$ 20' = -512$^{\circ}$ + $\frac{20}{60}$ -512$^{\circ}$ $\frac{20}{60}$ = -512$^{\circ}$ + 0.33$^{\circ}$ -312$^{\circ}$ $\frac{20}{60}$ = -512.33$^{\circ}$ $cot$(-512.33$^{\circ}$) $\approx$ 1.90714493 Therefore: $\cot$(-512$^{\circ}$ 20') $\approx$ 1.90714493