Question

question 13 (1 point) saved page 13 of 35 △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 = 13/12, sec r = 13/5, cot r = 12/5 b) csc r = 13/5, sec r = 13/12, cot r = 5/12 c) csc r = 12/13, sec r = 5/13, cot r = 12/5 d) csc r = 13/12, sec r = 13/5, cot r = 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 \(\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}\)), \(\csc R=\frac{26}{10}=\frac{13}{5}\)
• \(\sec R=\frac{q}{r}\) (since \(\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}\)), \(\sec R=\frac{26}{24}=\frac{13}{12}\)
• \(\cot R=\frac{r}{p}\) (since \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\)), \(\cot R=\frac{24}{10}=\frac{12}{5}\)

Answer:

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