Question

right triangle pqr is shown. what is the value of tan(q)? enter your answer in the box. tan(q) = \boxed{}

Explanation:

Step1: Recall tangent definition

In a right triangle, $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ to angle $\theta$.

Step2: Identify sides for $\angle Q$

For $\angle Q$, opposite side is $PR = 15$, adjacent side is $QR = 8$? Wait, no: Wait, right triangle at $R$, so $\angle R = 90^\circ$. So vertices: $P$, $Q$, $R$ with right angle at $R$. So angle at $Q$: sides: opposite to $Q$ is $PR$ (length 15), adjacent to $Q$ is $QR$ (length 8)? Wait no, wait: in triangle $PQR$, right-angled at $R$. So sides: $PR = 15$ (horizontal), $QR = 8$ (vertical), hypotenuse $PQ = 17$. So angle at $Q$: the sides: the side opposite to $Q$ is $PR$ (15), and the side adjacent to $Q$ is $QR$ (8)? Wait, no, wait: angle at $Q$: the two legs: one is $QR$ (from $Q$ to $R$, length 8), and the other is $QP$? No, wait, in angle $Q$, the sides forming the angle are $QQ$? No, angle at $Q$: the sides are $QP$ (hypotenuse, 17) and $QR$ (leg, 8), and the other leg is $PR$ (15). Wait, no: in angle $Q$, the adjacent side is the leg that is part of the angle, and the opposite side is the leg not part of the angle. So angle at $Q$: the sides: adjacent is $QR$ (length 8), opposite is $PR$ (length 15). Wait, no, wait: tangent of angle $Q$: in right triangle, $\tan(Q) = \frac{\text{opposite to } Q}{\text{adjacent to } Q}$. The opposite side to $Q$ is $PR$ (since $PR$ is opposite angle $Q$), and adjacent side is $QR$ (since $QR$ is adjacent to angle $Q$). Wait, but let's confirm: in triangle $PQR$, right-angled at $R$. So angle at $Q$: the sides: $QR$ is one leg (from $Q$ to $R$), $PR$ is the other leg (from $P$ to $R$), and $PQ$ is hypotenuse. So angle at $Q$: the side opposite is $PR$ (length 15), and the side adjacent is $QR$ (length 8). Wait, but wait, maybe I got it reversed. Wait, no: in angle $Q$, the angle is between $PQ$ (hypotenuse) and $QR$ (leg). So the adjacent side is $QR$ (length 8), and the opposite side is $PR$ (length 15). So $\tan(Q) = \frac{\text{opposite}}{\text{adjacent}} = \frac{PR}{QR} = \frac{15}{8}$? Wait, no, wait: wait, maybe I mixed up the angle. Wait, angle at $Q$: let's label the triangle: $P$---$R$ (15 units), $R$---$Q$ (8 units), right angle at $R$. So angle at $Q$: the angle between $Q$---$R$ (8) and $Q$---$P$ (17). So the side opposite to angle $Q$ is $P$---$R$ (15), and the side adjacent is $Q$---$R$ (8). So $\tan(Q) = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{8}$? Wait, but wait, maybe I made a mistake. Wait, no: in a right triangle, tangent of an angle is opposite over adjacent. So for angle $Q$, opposite side is $PR$ (15), adjacent is $QR$ (8). So $\tan(Q) = \frac{15}{8}$? Wait, but let's check with angle $P$. For angle $P$, opposite is $QR$ (8), adjacent is $PR$ (15), so $\tan(P) = \frac{8}{15}$. So for angle $Q$, it's $\frac{15}{8}$. Yes, that makes sense. So the steps:

1. Recall that in a right triangle, $\tan(\theta) = \frac{\text{opposite side to } \theta}{\text{adjacent side to } \theta}$.
2. For $\angle Q$ in right triangle $PQR$ (right-angled at $R$), the opposite side to $\angle Q$ is $PR = 15$, and the adjacent side is $QR = 8$.
3. Therefore, $\tan(Q) = \frac{PR}{QR} = \frac{15}{8}$.

Answer:

$\frac{15}{8}$