Question 1210134: In \triangle PQR, we have \angle P = 90^\circ, PQ=3, and QR = 5. Find \tan R. Found 2 solutions by CPhill, ikleyn:Answer by CPhill(1987) (Show Source):
You can put this solution on YOUR website! Let $\triangle PQR$ be a right triangle with $\angle P = 90^\circ$. We are given that $PQ = 3$ and $QR = 5$. We want to find $\tan R$.
Since $\triangle PQR$ is a right triangle, we can use the Pythagorean theorem to find the length of $PR$.
$$PQ^2 + PR^2 = QR^2$$
$$3^2 + PR^2 = 5^2$$
$$9 + PR^2 = 25$$
$$PR^2 = 25 - 9 = 16$$
$$PR = \sqrt{16} = 4$$
Now we can find $\tan R$. Recall that in a right triangle,
$$ \tan R = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{PQ}{PR} $$
We have $PQ = 3$ and $PR = 4$.
$$ \tan R = \frac{3}{4} $$
Final Answer: The final answer is $\boxed{\frac{3}{4}}$
