Question

question 13 (1 point)
△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 \( r=\sqrt{q^{2}-p^{2}}=\sqrt{26^{2}-10^{2}}=\sqrt{(26 + 10)(26-10)}=\sqrt{36\times16}=\sqrt{576}=24\)

Step2: Recall the reciprocal trigonometric ratios

• \(\csc R=\frac{q}{p}\) (since \(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\) for angle \(R\), hypotenuse \(q = 26\), opposite to \(R\) is \(p = 10\))
• \(\sec R=\frac{q}{r}\) (since \(\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\) for angle \(R\), hypotenuse \(q = 26\), adjacent to \(R\) is \(r = 24\))
• \(\cot R=\frac{r}{p}\) (since \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\) for angle \(R\), adjacent \(r = 24\), opposite \(p = 10\))

Substituting the values:

• \(\csc R=\frac{26}{24}=\frac{13}{12}\)
• \(\sec R=\frac{26}{10}=\frac{13}{5}\)
• \(\cot R=\frac{24}{10}=\frac{12}{5}\)

Answer:

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