Question

find the cotangent of $\angle p$.
simplify your answer and write it as a proper fraction, improper fraction, or whole number.
$\cot(p) = \square$

Explanation:

Step1: Recall cotangent definition

In a right triangle, \(\cot(\theta) = \frac{\text{adjacent side to } \theta}{\text{opposite side to } \theta}\). For \(\angle P\) in right triangle \(PQR\) (right - angled at \(Q\)):
The adjacent side to \(\angle P\) is \(PQ = 8\), and the opposite side to \(\angle P\) is \(QR\). First, we need to find the length of \(QR\) using the Pythagorean theorem. The Pythagorean theorem states that in a right triangle \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse and \(a,b\) are the legs. Here, hypotenuse \(PR = 17\) and leg \(PQ = 8\), let \(QR=x\). Then \(8^{2}+x^{2}=17^{2}\).

Step2: Solve for \(QR\)

\[ \begin{align*} x^{2}&=17^{2}-8^{2}\\ x^{2}&=(17 + 8)(17 - 8)\\ x^{2}&=25\times9\\ x^{2}&=225\\ x& = 15 \end{align*} \]

Step3: Calculate \(\cot(P)\)

Now, using the definition of cotangent, for \(\angle P\), adjacent side \(=PQ = 8\), opposite side \(=QR=15\). So \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{8}{15}\)? Wait, no. Wait, in right triangle \(PQR\), right - angled at \(Q\), the sides: \(PQ\) is one leg, \(QR\) is the other leg, and \(PR\) is the hypotenuse. For \(\angle P\), the adjacent side is \(PQ\) and the opposite side is \(QR\)? Wait, no. Wait, the angle at \(P\): the sides: the side adjacent to \(\angle P\) is \(PQ\) (because it forms \(\angle P\) with the hypotenuse \(PR\)), and the side opposite to \(\angle P\) is \(QR\). Wait, but let's re - check. The right angle is at \(Q\), so:
\(PQ\) and \(QR\) are the legs, \(PR\) is the hypotenuse. For angle \(P\):
- Adjacent side: the side that is part of \(\angle P\) and is not the hypotenuse. So adjacent side to \(\angle P\) is \(PQ = 8\)? Wait, no, wait. Wait, in angle \(P\), the two sides forming the angle are \(PQ\) and \(PR\). So the adjacent side is \(PQ\) (length 8) and the opposite side is \(QR\) (length we found as 15). Wait, but \(\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}\). Wait, no, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\), so \(\cot(\theta)=\frac{\text{adjacent}}{\text{opposite}}\). Wait, but let's re - express:
Wait, in right triangle, for angle \(P\):
\(\tan(P)=\frac{\text{opposite}}{\text{adjacent}}=\frac{QR}{PQ}\), so \(\cot(P)=\frac{PQ}{QR}\). Wait, we found \(QR = 15\) and \(PQ = 8\)? Wait, no, wait the calculation of \(QR\):
\(PR = 17\), \(PQ = 8\), so \(QR=\sqrt{17^{2}-8^{2}}=\sqrt{289 - 64}=\sqrt{225}=15\). Then, for angle \(P\), the adjacent side (to angle \(P\)) is \(PQ = 8\)? No, wait, angle \(P\) is at vertex \(P\), so the sides:
- The side opposite to angle \(P\) is \(QR\) (since it is across from \(P\)), and the side adjacent to angle \(P\) is \(PQ\) (since it is one of the sides forming angle \(P\) along with the hypotenuse \(PR\)). Wait, no, actually, in standard notation, in triangle \(PQR\) right - angled at \(Q\):
- \(\angle P\): the sides:
- Opposite side: \(QR\)
- Adjacent side: \(PQ\)
- Hypotenuse: \(PR\)
So \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{PQ}{QR}\). Wait, \(PQ = 8\), \(QR = 15\), so \(\cot(P)=\frac{8}{15}\)? Wait, no, that can't be. Wait, maybe I mixed up adjacent and opposite. Wait, let's think again. The tangent of an angle in a right triangle is opposite over adjacent. So \(\tan(P)=\frac{QR}{PQ}\), so \(\cot(P)=\frac{PQ}{QR}\). But let's check the lengths again. \(PQ = 8\), \(PR = 17\), so \(QR=\sqrt{17^{2}-8^{2}} = 15\). So \(\cot(P)=\frac{PQ}{QR}=\frac{8}{15}\)? Wait, no, wait, maybe I had the adjacent and opposite reversed. Wait, angle \(P\): the side adjacent to \(P\) is the side that is between \(P\) and the right angle \(Q\), which is \(PQ\), and the side opposite is \(QR\). So yes, \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{PQ}{QR}=\frac{8}{15}\)? Wait, no, wait, no. Wait, in the right triangle, for angle \(P\), the adjacent side is the leg that is part of angle \(P\) (other than the hypotenuse), and the opposite side is the leg that is not part of angle \(P\). So angle \(P\) is formed by \(PQ\) and \(PR\). So the adjacent side is \(PQ\) (length 8), and the opposite side is \(QR\) (length 15). So \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{8}{15}\)? Wait, but that seems wrong. Wait, no, wait, let's take an example. If we have a right triangle with legs 3 and 4, hypotenuse 5. For the angle opposite to 3, \(\cot(\theta)=\frac{4}{3}\) (adjacent over opposite). So in our case, angle \(P\): the side opposite to \(P\) is \(QR = 15\), the side adjacent to \(P\) is \(PQ = 8\). So \(\cot(P)=\frac{8}{15}\)? Wait, no, wait, I think I made a mistake. Wait, \(PQ\) is length 8, \(PR\) is 17, \(QR\) is 15. So angle at \(P\): the sides:
- Opposite: \(QR = 15\)
- Adjacent: \(PQ = 8\)
So \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{8}{15}\)? Wait, but let's check with tangent. \(\tan(P)=\frac{15}{8}\), so \(\cot(P)=\frac{8}{15}\). Yes, that's correct.

Wait, no, wait, I think I had the adjacent and opposite reversed. Wait, in the right triangle, the angle at \(P\): the side adjacent to \(P\) is the side that is connected to \(P\) and the right angle \(Q\), which is \(PQ\) (length 8), and the side opposite to \(P\) is \(QR\) (length 15). So \(\cot(P)=\frac{\text{adjacent}}{\text{opposite}}=\frac{8}{15}\)? Wait, no, that can't be. Wait, no, the adjacent side to an angle in a right triangle is the leg that is part of the angle (other than the hypotenuse), and the opposite side is the leg that is not part of the angle. So for angle \(P\), the two sides forming the angle are \(PQ\) and \(PR\). So the adjacent side is \(PQ\) (length 8), and the opposite side is \(QR\) (length 15). So \(\cot(P)=\frac{PQ}{QR}=\frac{8}{15}\). Wait, but let's calculate using the definition of cotangent as \(\frac{\cos(P)}{\sin(P)}\). \(\cos(P)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{8}{17}\), \(\sin(P)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{15}{17}\). Then \(\cot(P)=\frac{\cos(P)}{\sin(P)}=\frac{\frac{8}{17}}{\frac{15}{17}}=\frac{8}{15}\). Yes, that's correct.

Answer:

\(\frac{8}{15}\)