Question

find the value of cos p rounded to the nearest hundredth, if necessary.

(image of a right triangle pqr with right angle at q, pq = 15, pr = 28)

answer attempt 1 out of 2
cos p = |
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Explanation:

Step1: Recall cosine definition

In a right triangle, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ for angle $\theta$.

Step2: Identify sides for $\angle P$

For $\angle P$, adjacent side is $PQ = 15$? Wait, no. Wait, triangle $PQR$ is right-angled at $Q$. So sides: $PQ = 15$, $PR = 28$? Wait, no, $PR$ is the hypotenuse? Wait, no, right angle at $Q$, so legs are $PQ$ and $QR$, hypotenuse is $PR$. Wait, $PQ = 15$, $PR = 28$? Wait, no, $PR$ is one of the sides. Wait, let's label: right angle at $Q$, so vertices $P$, $Q$, $R$ with right angle at $Q$. So $PQ$ and $QR$ are legs, $PR$ is hypotenuse. Wait, given $PQ = 15$, $PR = 28$? Wait, no, $PR$ is length 28? Wait, no, the side from $P$ to $R$ is 28, and from $P$ to $Q$ is 15. Wait, for angle $P$, the adjacent side is $PQ$? No, adjacent side to angle $P$ is the leg that is part of angle $P$, which is $PQ$? Wait, no, angle $P$ is at vertex $P$, so the sides: the sides forming angle $P$ are $PQ$ (length 15) and $PR$ (length 28)? Wait, no, $PR$ is the side opposite to $Q$? Wait, no, right angle at $Q$, so $PQ$ and $QR$ are legs, $PR$ is hypotenuse. Wait, $PQ = 15$, $PR = 28$? Wait, no, $PR$ is the hypotenuse? Wait, no, hypotenuse is the side opposite the right angle, so $PR$ is hypotenuse. Wait, then for angle $P$, the adjacent side is $PQ$ (length 15) and the hypotenuse is $PR$ (length 28)? Wait, no, that can't be, because in a right triangle, hypotenuse is the longest side. Wait, 28 is longer than 15, so $PR = 28$ is hypotenuse, $PQ = 15$ is one leg, and $QR$ is the other leg. Wait, but for angle $P$, the adjacent side is $PQ$ (length 15) and the hypotenuse is $PR$ (length 28)? Wait, no, adjacent side to angle $P$ is the leg that is adjacent to angle $P$, which is $PQ$ (since angle $P$ is between $PQ$ and $PR$). Wait, no, angle $P$ is at vertex $P$, so the sides: from $P$ to $Q$ is 15, from $P$ to $R$ is 28, and from $Q$ to $R$ is the other leg. Wait, but cosine of angle $P$ is adjacent over hypotenuse. Adjacent side to angle $P$ is $PQ$ (length 15), hypotenuse is $PR$ (length 28)? Wait, no, that would make $\cos P = \frac{15}{28}$? But wait, 15 and 28: wait, no, maybe I mixed up. Wait, no, in a right triangle, cosine of an angle is adjacent over hypotenuse. So angle $P$: the sides: the leg adjacent to $P$ is $PQ$ (length 15), and the hypotenuse is $PR$ (length 28). Wait, but 15 is adjacent, 28 is hypotenuse? Wait, but 28 is longer than 15, so hypotenuse is 28. So $\cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{15}{28}$? Wait, no, that can't be, because 15/28 is approximately 0.5357, but wait, maybe I got the sides wrong. Wait, no, wait, the side from $P$ to $R$ is 28, which is one of the sides, and $PQ$ is 15. Wait, maybe $PR$ is not the hypotenuse. Wait, right angle at $Q$, so hypotenuse is $PR$, so $PR$ must be the longest side. 28 is longer than 15, so that's correct. So $PR = 28$ (hypotenuse), $PQ = 15$ (adjacent to angle $P$), so $\cos P = \frac{PQ}{PR} = \frac{15}{28} \approx 0.5357$, which rounds to 0.54? Wait, no, wait, 15 divided by 28: 15 ÷ 28 ≈ 0.5357, which is approximately 0.54 when rounded to the nearest hundredth. Wait, but wait, maybe I made a mistake in identifying adjacent side. Wait, angle $P$: the sides: the two sides forming angle $P$ are $PQ$ (15) and $PR$ (28)? No, angle $P$ is at $P$, so the sides are $PQ$ (from $P$ to $Q$) and $PR$ (from $P$ to $R$), with the right angle at $Q$. So the adjacent side to angle $P$ is $PQ$ (length 15), and the hypotenuse is $PR$ (length 28). So $\cos P = \frac{15}{28} \approx 0.54$ (rounded to nearest hundredth). Wait, but let's check: 15 divided by 28. 28 × 0.5 = 14, 28 × 0.53 = 14.84, 28 × 0.54 = 15.12. So 15 is between 14.84 and 15.12. 15 - 14.84 = 0.16, 15.12 - 15 = 0.12. So closer to 0.54? Wait, 15/28 ≈ 0.5357, which is 0.54 when rounded to nearest hundredth.

Step1: Correct identification

Wait, no, wait a second. Maybe I mixed up the sides. Let's re-express the triangle: right-angled at $Q$, so $PQ$ and $QR$ are legs, $PR$ is hypotenuse. Given $PQ = 15$, $PR = 28$? Wait, no, $PR$ is length 28, so hypotenuse is 28. Then for angle $P$, the adjacent side is $PQ$ (length 15), and the hypotenuse is $PR$ (length 28). So $\cos P = \frac{PQ}{PR} = \frac{15}{28} \approx 0.54$ (rounded to nearest hundredth). Wait, but 15/28 is approximately 0.5357, which is 0.54 when rounded to the nearest hundredth.

Step2: Calculate

$\cos P = \frac{15}{28} \approx 0.54$ (rounded to nearest hundredth). Wait, no, 15 divided by 28: 15 ÷ 28 ≈ 0.5357, which is 0.54 when rounded to the nearest hundredth (since the third decimal is 5, we round up the second decimal: 0.5357 ≈ 0.54).

Answer:

$\boxed{0.54}$