Question

question 13 (1 point) ✓ saved

△pqr has a right angle at q. if q = 26 cm and p = 10 cm, which of the following is the correct set of reciprocal trigonometric ratios for ∠r?

a)

csc r = \\( \frac { 13 } { 12 } \\), sec r = \\( \frac { 13 } { 5 } \\),

cot r = \\( \frac { 12 } { 5 } \\)

b)

csc r = \\( \frac { 13 } { 5 } \\), sec r = \\( \frac { 13 } { 12 } \\),

cot r = \\( \frac { 5 } { 12 } \\)

c)

csc r = \\( \frac { 12 } { 13 } \\), sec r = \\( \frac { 5 } { 13 } \\),

cot r = \\( \frac { 12 } { 5 } \\)

d)

csc r = \\( \frac { 13 } { 12 } \\), sec r = \\( \frac { 13 } { 5 } \\),

cot r = \\( \frac { 5 } { 12 } \\)

Explanation:

Step1: Find the length of side \( r \)

Using the Pythagorean theorem \( q^{2}=p^{2}+r^{2} \). Given \( q = 26 \) and \( p = 10 \), we have \( 26^{2}=10^{2}+r^{2} \), so \( r^{2}=26^{2}-10^{2}=(26 + 10)(26 - 10)=36\times16 \), and \( r = 24 \).

Step2: Recall reciprocal trigonometric ratios

\(\csc R=\frac{q}{p}\), \(\sec R=\frac{q}{r}\), \(\cot R=\frac{r}{p}\)
Since \( q = 26 \), \( p = 10 \), \( r = 24 \)
\(\csc R=\frac{26}{10}=\frac{13}{5}\), \(\sec R=\frac{26}{24}=\frac{13}{12}\), \(\cot R=\frac{24}{10}=\frac{12}{5}\)

Answer:

B. \(\csc R=\frac{13}{5},\sec R=\frac{13}{12},\cot R=\frac{12}{5}\)