<Question>
In $\triangle{RST}$, shown, $\sin{R}=\frac{2}{5}$.  What is $\sin{T}$?

[asy]
pair R,S,T;
T = (0,0);
S = (2,0);
R = (2,sqrt(21));
draw(R--S--T--R);
draw(rightanglemark(R,S,T,10));
label("$T$",T,SW);
label("$S$",S,SE);
label("$R$",R,NE);
label("$5$",(R+T)/2,NW);
[/asy]
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find sin T in triangle RST. The problem says that sin R is 2/5. There's also an Asymptote figure provided, which I can't see, but the code mentions that angle S is a right angle because it draws the right angle mark between R, S, and T. So, triangle RST is a right triangle with the right angle at S. Let me confirm that: the coordinates given are T at (0,0), S at (2,0), and R at (2, sqrt(21)). So, connecting R(2, sqrt(21)) to S(2,0) is a vertical line segment, and S(2,0) to T(0,0) is a horizontal line segment. Then R to T would be the hypotenuse. So, the right angle is at S. Therefore, triangle RST is a right-angled triangle at S.

Given that, let's recall that in a right-angled triangle, the sum of the other two angles is 90 degrees, so angles R and T must add up to 90 degrees. That might come in handy.

The problem states sin R = 2/5. Since angle R is one of the acute angles in the right triangle, sin R would be the ratio of the opposite side to the hypotenuse. Wait, in triangle RST, angle R is at point R. Let me make sure: in triangle RST, the vertices are R, S, T. Since the right angle is at S, the sides opposite to the angles would be as follows: side opposite angle R is ST, side opposite angle T is RS, and the hypotenuse is RT.

<prompt_1.1> Identify the sides of the triangle opposite to angles R and T, and express sin R and sin T in terms of these sides. </prompt_1.1>
-----

But let me verify this with the given information. The problem says sin R is 2/5. Let's check that with the sides.

<prompt_1.2> Verify the calculation of sin T using the given value of sin R and the relationship between angles R and T. </prompt_1.2>

Alternatively, since angles R and T are complementary, sin T = cos R. Because in a right triangle, sin(theta) = cos(90 - theta). Since angle T is 90 - angle R, sin T = cos R. So if we can find cos R, that would be sin T.

<prompt_1.3> Use the complementary angle relationship between R and T to find sin T as cos R. </prompt_1.3>

Alternatively, if I use the side ratios. Let's see. Let's try to compute the sides. Let's denote the sides as follows:

<prompt_1.4> Explore alternative methods to find sin T using different side ratios and trigonometric identities. </prompt_1.4>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, no. Wait, in a triangle, the side opposite an angle is named accordingly. So, in triangle RST, angle R is at vertex R, so the side opposite angle R is ST. Similarly, the side opposite angle T is RS. The hypotenuse is RT, since it's opposite the right angle at S.

So, sin R would be opposite over hypotenuse, which is ST / RT. Similarly, sin T would be opposite over hypotenuse, which is RS / RT.
</execution_1.1>

<execution_1.2>
In triangle RST, angle R is at point R. The side opposite angle R is ST. The hypotenuse is RT. So sin R = opposite/hypotenuse = ST / RT. According to the problem, sin R is 2/5. Therefore, ST / RT = 2/5. So ST = (2/5) * RT.

But we can also find the lengths of the sides from the coordinates given. Let me check the coordinates again. T is at (0,0), S is at (2,0), and R is at (2, sqrt(21)). So, the length of TS is the distance from T(0,0) to S(2,0), which is 2 units. The length of SR is the distance from S(2,0) to R(2, sqrt(21)), which is sqrt(21) units. The hypotenuse RT is the distance from R(2, sqrt(21)) to T(0,0). Let's compute that: sqrt[(2-0)^2 + (sqrt(21)-0)^2] = sqrt[4 + 21] = sqrt[25] = 5. So RT is 5 units. Therefore, ST is 2 units, RT is 5 units, and SR is sqrt(21) units.

Therefore, sin R is opposite over hypotenuse, which is ST / RT = 2/5, which matches the given information.
</execution_1.2>

<execution_1.3>
Given sin R = 2/5, then cos R would be sqrt(1 - sin^2 R) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5. Therefore, sin T = sqrt(21)/5.
</execution_1.3>

<execution_1.4>
- Let's call side opposite angle R as ST. Since angle R is at R, the opposite side is ST.
- Side opposite angle T is RS.
- Hypotenuse is RT.

Given that, sin R = opposite / hypotenuse = ST / RT = 2/5. So ST = 2 and RT = 5. Then, by Pythagoras, RS^2 + ST^2 = RT^2. Wait, but RS is one leg, ST is another leg, and RT is the hypotenuse. Wait, no: in a right-angled triangle at S, the legs are RS and ST, and the hypotenuse is RT. So, yes, RS^2 + ST^2 = RT^2. So if ST is 2 and RT is 5, then RS^2 + 2^2 = 5^2 => RS^2 + 4 = 25 => RS^2 = 21 => RS = sqrt(21).
</execution_1.4>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But we don't know the lengths of the sides yet. However, the Asymptote code gives coordinates, so maybe we can compute the sides from the coordinates? Let me check.

<prompt_2.1> Calculate the lengths of the sides RT, RS, and ST using the given coordinates. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
The coordinates are:

- T is at (0,0)

- S is at (2,0)

- R is at (2, sqrt(21))

So, triangle RST has vertices at these three points. Let's compute the lengths of the sides.

First, side RS: from R(2, sqrt(21)) to S(2,0). Since the x-coordinates are the same, it's a vertical line. The length is the difference in y-coordinates: sqrt(21) - 0 = sqrt(21). So RS = sqrt(21).

Side ST: from S(2,0) to T(0,0). The y-coordinates are the same, so it's a horizontal line. The length is the difference in x-coordinates: 2 - 0 = 2. So ST = 2.

Hypotenuse RT: from R(2, sqrt(21)) to T(0,0). The distance formula: sqrt[(2-0)^2 + (sqrt(21)-0)^2] = sqrt[4 + 21] = sqrt[25] = 5. So RT = 5.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Therefore, side opposite angle R is ST = 2, hypotenuse RT = 5, so sin R = 2/5, which matches the given information.
Therefore, side opposite angle T is RS = sqrt(21), so sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5.

Wait, but let me make sure. Let's confirm the sides again.

<prompt_3.1> Confirm the calculation of sin T using the side lengths and the relationship between angles R and T. </prompt_3.1>
-----

Based on execution_2.1, execution_1.3:
Therefore, sin T = sqrt(21)/5.

Alternatively, if I use angle T, which is at point T(0,0). The sides around angle T: in the right triangle, angle T is between sides TS and TT (but TT is the hypotenuse). Wait, no. Wait, angle T is at point T, which is connected to S and R. So, the sides forming angle T are TS and TR. Wait, no, in the triangle RST, angle T is between sides TS and TR. Wait, perhaps it's better to stick with the side opposite over hypotenuse.

<prompt_3.2> Reconfirm the calculation of sin T using the side lengths and trigonometric identities. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
In triangle RST, right-angled at S. So:

- Angle R is at R(2, sqrt(21)). The sides adjacent to angle R would be RS and the hypotenuse RT. Wait, no. Wait, in a right-angled triangle, each angle's sine is opposite over hypotenuse. For angle R, the opposite side is ST, and the hypotenuse is RT. For angle T, the opposite side is RS, and the hypotenuse is RT. Therefore, sin T = RS / RT = sqrt(21)/5.
</execution_3.1>

<execution_3.2>
Since angle T is at point T, the side opposite angle T is RS. Because in triangle RST, the side opposite angle T is the side that doesn't touch angle T, which is RS. Therefore, sin T = opposite / hypotenuse = RS / RT = sqrt(21)/5.

Alternatively, using the coordinates, we can compute angle T. Since angle T is at the origin (0,0), between points S(2,0), T(0,0), and R(2, sqrt(21)). So, angle T is the angle between the x-axis (from T to S) and the line from T to R, which goes up to (2, sqrt(21)). The coordinates of R relative to T are (2, sqrt(21)), so the slope of TR is (sqrt(21)/2). Therefore, the tangent of angle T is opposite over adjacent, which is sqrt(21)/2. Therefore, tan T = sqrt(21)/2. Then, sin T can be found from tan T. If tan T = opposite/adjacent = sqrt(21)/2, then we can imagine a right triangle with opposite side sqrt(21), adjacent side 2, hypotenuse sqrt( (sqrt(21))^2 + 2^2 ) = sqrt(21 + 4) = sqrt(25) = 5. Therefore, sin T = opposite / hypotenuse = sqrt(21)/5.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Therefore, the answer should be sqrt(21)/5. Let me check if that's in the simplest form.
The numerator is sqrt(21), denominator is 5. 21 has no square factors, so yes, that's simplified.

Alternatively, we can use the fact that in a right-angled triangle, the sines of the two acute angles are equal to the adjacent over hypotenuse for the other angle. Wait, maybe that's complicating.

<prompt_4.1> Perform a final verification of the solution using trigonometric identities and relationships between angles. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
But given that angle R and angle T are complementary, sin T = cos R. Since sin R = 2/5, then cos R = sqrt(1 - (2/5)^2) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5, which is the same as sin T. So that also confirms it.

Therefore, regardless of the method used—direct computation from coordinates, using trigonometric identities, or the complementary angle relationship—the value of sin T is sqrt(21)/5.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
**Final Answer**
\boxed{\dfrac{\sqrt{21}}{5}}
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, $\sin{T} = \frac{\sqrt{21}}{5}$, which can be written as $\boxed{\frac{\sqrt{21}}{5}}$.
</Final_answer>
</think>