Chapter 2 - Acute Angles and Right Triangles - Section 2.1 Trigonometric Functions of Acute Angles - 2.1 Exercises: 56

$\sec$ 45$^{\circ}$ = $\sqrt2$

Work Step by Step

$\sec$ 45$^{\circ}$ We must find side $x$ of the 45$^{\circ}$ - 45$^{\circ}$ Right Triangle. Pythagorean Theorem: $c$$^{2} = a$$^{2}$ + $b$$^{2}$ x$^{2}$ = 1$^{2}$ + 1$^{2}$ x$^{2}$ = 2 x = $\sqrt2$ Now consider the triangle from the perspective of a 45$^{\circ}$ angle. Hypotenuse = $\sqrt2$ Opposite = 1 Adjacent = 1 $\csc$ 45$^{\circ}$ = $\frac{Hypotenuse}{Adjacent}$ = $\frac{\sqrt2}{1}$ = $\sqrt2$ Therefore: $\sec$ 45$^{\circ}$ = $\sqrt2$

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