Question

what is sec(∠b)? reduce fractional answers to lowest terms.

Explanation:

Step1: Recall secant definition

In a right triangle, \(\sec(\theta)=\frac{\text{Hypotenuse}}{\text{Adjacent side to } \theta}\). For \(\angle B\), identify hypotenuse and adjacent side.
Hypotenuse \(AB = 10\), adjacent side to \(\angle B\) is \(BC = 6\)? Wait, no—wait, right triangle \( \triangle ABC\) with right angle at \(C\). So \(\angle B\): adjacent side is \(BC\)? Wait, no, adjacent side to \(\angle B\) is the leg adjacent, which is \(BC\)? Wait, no, in right triangle, for angle \(B\), the sides: opposite is \(AC = 8\), adjacent is \(BC = 6\), hypotenuse \(AB = 10\). Wait, but \(\sec(\angle B)=\frac{1}{\cos(\angle B)}\), and \(\cos(\angle B)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{BC}{AB}\)? Wait no, adjacent to \(\angle B\) is \(BC\)? Wait, angle \(B\) is at vertex \(B\), so the sides: the two legs are \(BC\) (length 6) and \(AC\) (length 8), hypotenuse \(AB\) (length 10). So for angle \(B\), the adjacent side is \(BC\) (adjacent to angle \(B\), since it's one of the legs forming angle \(B\) with hypotenuse), and opposite side is \(AC\). Wait, no: in angle \(B\), the sides: the adjacent side is the leg that is part of angle \(B\) (other than hypotenuse), so \(BC\) is adjacent? Wait, no, angle \(B\) is between \(AB\) (hypotenuse) and \(BC\) (leg), so adjacent side is \(BC\), hypotenuse is \(AB\). Wait, but \(\cos(\angle B)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{BC}{AB}=\frac{6}{10}\)? Wait, no, that can't be. Wait, no, adjacent side to angle \(B\) is \(BC\)? Wait, no, let's label the triangle: right angle at \(C\), so \(AC\) and \(BC\) are legs, \(AB\) hypotenuse. So angle at \(B\): the sides: adjacent is \(BC\) (length 6), opposite is \(AC\) (length 8), hypotenuse \(AB\) (length 10). Then \(\cos(\angle B)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{BC}{AB}=\frac{6}{10}\)? Wait, no, that's incorrect. Wait, no: adjacent side to angle \(B\) is the leg that is adjacent to angle \(B\), which is \(BC\)? Wait, no, angle \(B\) is formed by \(AB\) (hypotenuse) and \(BC\) (leg), so the adjacent side is \(BC\), and the opposite side is \(AC\). But then \(\cos(\angle B)=\frac{BC}{AB}=\frac{6}{10}\), so \(\sec(\angle B)=\frac{1}{\cos(\angle B)}=\frac{AB}{BC}=\frac{10}{6}=\frac{5}{3}\). Wait, let's check again. Wait, maybe I mixed up adjacent and opposite. Wait, in angle \(B\), the adjacent side is the side that is not the hypotenuse and is next to angle \(B\). So angle \(B\) is at vertex \(B\), so the two sides meeting at \(B\) are \(AB\) (hypotenuse) and \(BC\) (leg). So the adjacent side is \(BC\) (length 6), and the opposite side is \(AC\) (length 8). Then \(\cos(\angle B)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{BC}{AB}=\frac{6}{10}\), so \(\sec(\angle B)=\frac{1}{\cos(\angle B)}=\frac{AB}{BC}=\frac{10}{6}=\frac{5}{3}\). Yes, that makes sense.

Step2: Simplify the fraction

\(\frac{10}{6}\) simplifies by dividing numerator and denominator by 2: \(\frac{10\div2}{6\div2}=\frac{5}{3}\).

Answer:

\(\frac{5}{3}\)