Question

find sinθ, where θ is the angle shown. give an exact value, not a decimal approximation.

Explanation:

Step1: Recall sine definition

In a right triangle, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$. First, find the hypotenuse using Pythagoras: $c = \sqrt{a^2 + b^2}$.
Here, $a = 3$, $b = 5$, so $c = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$.

Step2: Apply sine formula

Opposite side to $\theta$ is $5$, hypotenuse is $\sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}}$. Rationalize: $\frac{5\sqrt{34}}{34}$. Wait, no—wait, opposite? Wait, $\theta$ is at the base angle. Wait, adjacent is 3, opposite is 5? Wait, no: in the right triangle, the right angle is between 3 and 5. So $\theta$ is at the vertex with adjacent 3, opposite 5, hypotenuse $\sqrt{3^2 + 5^2} = \sqrt{34}$. So $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{\sqrt{34}}$? Wait, no—wait, no: wait, the angle $\theta$: the side opposite $\theta$ is 5? Wait, no, the vertical side is 5, horizontal is 3. So $\theta$ is at the bottom left, so opposite side is the vertical side (5), adjacent is horizontal (3), hypotenuse is $\sqrt{3^2 + 5^2} = \sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}}$? Wait, no, wait: $\sin\theta$ is opposite over hypotenuse. Wait, no, maybe I mixed up. Wait, no: in a right triangle, for angle $\theta$, opposite is the side not adjacent to $\theta$. So if $\theta$ is at the bottom, the opposite side is the vertical leg (length 5), adjacent is horizontal (3), hypotenuse $\sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}}$? Wait, but let's check again. Wait, Pythagoras: $3^2 + 5^2 = 9 + 25 = 34$, so hypotenuse is $\sqrt{34}$. Then $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{\sqrt{34}}$, but rationalized, it's $\frac{5\sqrt{34}}{34}$. Wait, but maybe I made a mistake. Wait, no—wait, the problem: the triangle has legs 3 and 5, right angle between them. So $\theta$ is at the end of the 3 side. So opposite side to $\theta$ is 5, adjacent is 3, hypotenuse $\sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}}$ or rationalized. Wait, but maybe the problem expects the unrationalized? Wait, no, let's confirm. Wait, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$. Opposite is 5, hypotenuse $\sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}} = \frac{5\sqrt{34}}{34}$. Wait, but maybe I messed up opposite and adjacent. Wait, no: $\theta$ is at the vertex with the horizontal side 3. So the side opposite $\theta$ is the vertical side (5), adjacent is 3. So yes, $\sin\theta = \frac{5}{\sqrt{34}}$, which rationalizes to $\frac{5\sqrt{34}}{34}$. Wait, but let's check again. Wait, maybe the hypotenuse is $\sqrt{3^2 + 5^2} = \sqrt{34}$, correct. So $\sin\theta = \frac{5}{\sqrt{34}}$ or $\frac{5\sqrt{34}}{34}$. But maybe the problem allows the denominator with radical, but usually rationalized. Wait, but let's see: the problem says "exact value, not a decimal". So either form is exact, but rationalized is preferred. Wait, but maybe I made a mistake in opposite/adjacent. Wait, no: $\sin\theta$ is opposite over hypotenuse. So if $\theta$ is at the bottom, opposite is up (5), hypotenuse is $\sqrt{34}$. So yes, $\sin\theta = \frac{5}{\sqrt{34}} = \frac{5\sqrt{34}}{34}$. Wait, but let's check with the triangle. Wait, maybe the vertical side is 5, horizontal 3, so hypotenuse $\sqrt{34}$. So $\sin\theta = \frac{5}{\sqrt{34}}$. Alternatively, maybe I had opposite and adjacent reversed. Wait, no: $\cos\theta$ would be adjacent over hypotenuse, so $\cos\theta = \frac{3}{\sqrt{34}}$, $\sin\theta = \frac{5}{\sqrt{34}}$. Yes, that's correct. So the exact value is $\frac{5\sqrt{34}}{34}$ or $\frac{5}{\sqrt{34}}$. But rationalized is standard. So $\sin\theta = \frac{5\sqrt{34}}{34}$. Wait, but let's check the calculation again. Wait, 3 squared is 9, 5 squared is 25, sum is 34, square root of 34. Opposite side to $\theta$ is 5, so $\sin\theta = 5 / \sqrt{34} = 5\sqrt{34}/34$. Yes.

Wait, wait, no—wait a second. Wait, the triangle: the right angle is between 3 and 5, so the two legs are 3 and 5, hypotenuse $\sqrt{3^2 + 5^2} = \sqrt{34}$. The angle $\theta$ is at the vertex where the leg 3 is adjacent, so the opposite leg is 5. So $\sin\theta = \text{opposite}/\text{hypotenuse} = 5/\sqrt{34}$, which is equal to $5\sqrt{34}/34$ when rationalized. So that's the exact value.

Answer:

$\frac{5\sqrt{34}}{34}$ (or $\frac{5}{\sqrt{34}}$ if not rationalized, but rationalized is standard)