question 10 (1 point)
riangle 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 angle r?
a)
csc r = \frac{13}{5}, sec r = \frac{13}{12}, cot r = \frac{5}{12}
b)
csc r = \frac{13}{12}, sec r = \frac{13}{5}, cot r = \frac{12}{5}
c)
csc r = \frac{12}{13}, sec r = \frac{5}{13}, cot r = \frac{12}{5}
d)
csc r = \frac{12}{13}, sec r = \frac{13}{5}, cot r = \frac{12}{5}

Question

question 10 (1 point)
	riangle 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 angle r?
	 a)
	csc r = \frac{13}{5}, sec r = \frac{13}{12}, cot r = \frac{5}{12}
	 b)
	csc r = \frac{13}{12}, sec r = \frac{13}{5}, cot r = \frac{12}{5}
	 c)
	csc r = \frac{12}{13}, sec r = \frac{5}{13}, cot r = \frac{12}{5}
	 d)
	csc r = \frac{12}{13}, sec r = \frac{13}{5}, cot r = \frac{12}{5}

Accepted Answer

# Explanation:
## Step1: Find the length of side $ r $
Using the Pythagorean theorem $ q^{2}=p^{2}+r^{2} $, where $ q = 26 $ and $ p = 10 $.
$$
\begin{align*}
r^{2}&=q^{2}-p^{2}\
r^{2}&=26^{2}-10^{2}\
r^{2}&=(26 + 10)(26-10)\
r^{2}&=36\times16\
r^{2}&=576\
r& = 24
\end{align*}
$$
## Step2: Calculate $ \csc R $ and $ \sec R $ and $ \cot R $
Recall the 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:
B. $ \csc R=\frac{13}{5},\sec R=\frac{13}{12},\cot R=\frac{12}{5}$

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QUESTION IMAGE

Question

Explanation:

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

Step2: Calculate \( \csc R \) and \( \sec R \) and \( \cot R \)

Answer: