Question

find pr.
write your answer as an integer or as a decimal rounded to the nearest tenth.
pr =
submit

Explanation:

Step1: Identify the trigonometric ratio

In right triangle \( PQR \) with \( \angle Q = 90^\circ \), \( \angle R = 27^\circ \), and \( PQ = 6 \) (adjacent to \( \angle R \)), \( PR \) is the hypotenuse. We use the cosine function: \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(27^\circ)=\frac{QR}{PR} \)? Wait, no, \( QR = 6 \) is adjacent to \( \angle R \), and \( PR \) is the hypotenuse. Wait, \( \angle R = 27^\circ \), \( QR = 6 \) (adjacent side to \( \angle R \)), so \( \cos(27^\circ)=\frac{QR}{PR} \), so \( PR=\frac{QR}{\cos(27^\circ)} \). Wait, no, \( QR \) is adjacent, \( PR \) is hypotenuse, so \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(27^\circ)=\frac{QR}{PR} \), so \( PR = \frac{QR}{\cos(27^\circ)} \). Wait, \( QR = 6 \), so \( PR=\frac{6}{\cos(27^\circ)} \). Let's calculate \( \cos(27^\circ) \approx 0.8910 \), so \( PR \approx \frac{6}{0.8910} \approx 6.73 \)? Wait, no, maybe I mixed up. Wait, \( \angle R = 27^\circ \), \( PQ \) is opposite to \( \angle R \), \( QR \) is adjacent, \( PR \) is hypotenuse. Wait, the right angle is at \( Q \), so \( PQ \perp QR \). So \( \angle Q = 90^\circ \), \( \angle R = 27^\circ \), \( QR = 6 \) (adjacent to \( \angle R \)), \( PR \) is hypotenuse. So \( \cos(27^\circ)=\frac{QR}{PR} \), so \( PR = \frac{QR}{\cos(27^\circ)} \). Let's compute \( \cos(27^\circ) \approx 0.8910 \), so \( PR \approx \frac{6}{0.8910} \approx 6.73 \)? Wait, no, maybe I made a mistake. Wait, maybe it's \( \cos(27^\circ)=\frac{QR}{PR} \), so \( PR = \frac{6}{\cos(27^\circ)} \approx \frac{6}{0.8910} \approx 6.73 \), but let's check with calculator. Alternatively, maybe \( \angle R = 27^\circ \), \( PQ \) is opposite, \( QR \) is adjacent, \( PR \) is hypotenuse. So \( \cos(27^\circ) = \frac{QR}{PR} \), so \( PR = \frac{QR}{\cos(27^\circ)} \approx \frac{6}{0.8910} \approx 6.73 \), but maybe I should use \( \cos(27^\circ) \approx 0.8910 \), so \( 6 / 0.8910 \approx 6.73 \), rounded to nearest tenth is 6.7? Wait, no, maybe I mixed up with sine. Wait, \( \sin(27^\circ)=\frac{PQ}{PR} \), \( \cos(27^\circ)=\frac{QR}{PR} \), \( \tan(27^\circ)=\frac{PQ}{QR} \). So if \( QR = 6 \), and we need \( PR \), then \( \cos(27^\circ)=\frac{6}{PR} \), so \( PR = \frac{6}{\cos(27^\circ)} \approx 6 / 0.8910 \approx 6.73 \), so rounded to nearest tenth is 6.7? Wait, but maybe I should use \( \cos(27^\circ) \approx 0.891 \), so \( 6 / 0.891 \approx 6.73 \), so 6.7 when rounded to nearest tenth. Wait, but let's check with calculator. Let me compute \( \cos(27^\circ) \): using calculator, \( \cos(27^\circ) \approx 0.891006524 \), so \( 6 / 0.891006524 \approx 6.734 \), so rounded to nearest tenth is 6.7. Wait, but maybe the problem is using \( \cos(27^\circ) \approx 0.891 \), so 6 / 0.891 ≈ 6.73, so 6.7. Wait, but let's confirm. Alternatively, maybe it's \( \cos(27^\circ) \approx 0.891 \), so \( PR \approx 6 / 0.891 \approx 6.7 \). Wait, but maybe I made a mistake in the trigonometric ratio. Let's re-express: in right triangle \( PQR \), right-angled at \( Q \), so:

- \( \angle Q = 90^\circ \)
- \( \angle R = 27^\circ \)
- \( QR = 6 \) (adjacent to \( \angle R \))
- \( PR \) is hypotenuse

So, \( \cos(\angle R) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{QR}{PR} \)

Therefore, \( PR = \frac{QR}{\cos(\angle R)} = \frac{6}{\cos(27^\circ)} \)

Calculating \( \cos(27^\circ) \approx 0.8910 \)

So, \( PR \approx \frac{6}{0.8910} \approx 6.73 \), which rounds to 6.7 when rounded to the nearest tenth. Wait, but let's check with a calculator. Let's compute \( 6 \div \cos(27^\circ) \):

\( \cos(27^\circ) \approx 0.891006524 \)

\( 6 \div 0.891006524 \approx 6.734 \)

Rounded to the nearest tenth is 6.7. Wait, but maybe the problem expects using \( \cos(27^\circ) \approx 0.891 \), so 6 / 0.891 ≈ 6.73, which is 6.7 when rounded to the nearest tenth. Alternatively, maybe I mixed up with adjacent and hypotenuse. Wait, maybe \( QR \) is adjacent, \( PR \) is hypotenuse, so yes, \( \cos(27^\circ) = \frac{QR}{PR} \), so \( PR = \frac{QR}{\cos(27^\circ)} \approx 6.7 \).

Step2: Calculate the value

Using a calculator, \( \cos(27^\circ) \approx 0.8910 \)

\( PR = \frac{6}{0.8910} \approx 6.73 \), which rounds to 6.7 when rounded to the nearest tenth. Wait, but let's check again. Wait, maybe the problem is using \( \cos(27^\circ) \approx 0.891 \), so \( 6 / 0.891 \approx 6.73 \), so 6.7. Alternatively, maybe I made a mistake in the trigonometric ratio. Let's check with sine. Wait, if \( \angle R = 27^\circ \), \( PQ \) is opposite, \( PR \) is hypotenuse, so \( \sin(27^\circ) = \frac{PQ}{PR} \), but we don't know \( PQ \). We know \( QR = 6 \), which is adjacent. So cosine is correct. So the calculation is \( PR = \frac{6}{\cos(27^\circ)} \approx 6.7 \) (rounded to nearest tenth).

Answer:

\( 6.7 \)