Okay, so I need to find sin T in triangle RST where angle R is given as 2/5. Let me start by visualizing the triangle. The Asymptote code shows that point T is at (0,0), S is at (2,0), and R is at (2, sqrt(21)). So, triangle RST is a right-angled triangle at S. Wait, hold on, the Asymptote code says rightanglemark(R,S,T,10), which means the right angle is at S. So, angle S is 90 degrees. That's important because in a right-angled triangle, we can use Pythagoras' theorem and trigonometric ratios more easily.

Given that sin R = 2/5. Let me recall that in a right-angled triangle, the sine of an angle is the ratio of the opposite side to the hypotenuse. So, in triangle RST, angle R is one of the acute angles. Since angle S is the right angle, the sides opposite angles R and T must be the legs of the triangle. Let me label the sides accordingly.

In triangle RST, right-angled at S, so side opposite angle R is ST, and the side opposite angle T is RS. The hypotenuse is RT. Wait, but in the Asymptote code, R is at (2, sqrt(21)), so the coordinates are T(0,0), S(2,0), R(2, sqrt(21)). So, RT is the hypotenuse, RS is the vertical side from R to S, which has length sqrt(21), and ST is the horizontal side from S to T, which has length 2.

Wait, but according to the Asymptote code, the hypotenuse is RT, which is between R(2, sqrt(21)) and T(0,0). Let me calculate the length of RT. Using the distance formula: sqrt[(2-0)^2 + (sqrt(21)-0)^2] = sqrt[4 + 21] = sqrt[25] = 5. So, the hypotenuse RT is 5 units, which matches the given information that the segment RT is labeled as 5. So, that's consistent.

Therefore, the sides are as follows: RT (hypotenuse) is 5, ST is 2, and RS is sqrt(21). Let me verify that using Pythagoras: 2^2 + (sqrt(21))^2 = 4 + 21 = 25, which is 5^2. Correct.

Now, angle R is at point R. Let me confirm that. The angles in a triangle are at each vertex. So, angle R is at point R, which is the vertex between sides RS and RT. Wait, but in the Asymptote code, the triangle is drawn as R--S--T--R, so the vertices are R, S, T. So, angle R is at point R between sides RS and RT. But in the right-angled triangle, angle S is the right angle, so angles R and T are the two acute angles.

Given that sin R = 2/5. Let me check that. In triangle RST, angle R is at point R. The sides adjacent to angle R are RS and RT. Wait, no. Wait, in a triangle, the sine of an angle is opposite over hypotenuse. So, for angle R, the opposite side is ST, which is 2, and the hypotenuse is RT, which is 5. Therefore, sin R = opposite/hypotenuse = ST/RT = 2/5. That's exactly what's given. So, that checks out.

Therefore, the given triangle has sides ST = 2, RS = sqrt(21), RT = 5. So, angle T is at point T, between sides ST and RT. The side opposite angle T is RS, which is sqrt(21). Therefore, sin T = opposite/hypotenuse = RS/RT = sqrt(21)/5. So, that's the answer. Wait, but let me confirm again.

In triangle RST, angle T is at point T, which is the right angle vertex. Wait, hold on. Wait, the triangle is right-angled at S, so angle S is 90 degrees. Therefore, angles R and T are the other two angles, both acute. So, angle T is at point T, which is the vertex opposite side RS. Wait, no, in the triangle, the side opposite angle T would be RS, which is sqrt(21). Therefore, sin T = opposite/hypotenuse = RS/RT = sqrt(21)/5. So, that's the answer. Therefore, sin T is sqrt(21)/5.

Wait, but maybe I should verify using another method. Let's use trigonometric identities. Since in a right-angled triangle, the two acute angles are complementary, meaning they add up to 90 degrees. So, angle R + angle T = 90 degrees. Therefore, sin T = cos R. Since sin R = 2/5, then cos R = sqrt(1 - sin² R) = sqrt(1 - 4/25) = sqrt(21/25) = sqrt(21)/5. Therefore, sin T = sqrt(21)/5. That matches the previous result. So, that's correct.

Alternatively, using the sides, since angle T is opposite side RS (length sqrt(21)) and hypotenuse RT (length 5), sin T = sqrt(21)/5. So, both methods confirm that the answer is sqrt(21)/5.

Therefore, the final answer is sqrt(21)/5, which can be written as √21⁄5. Since the question asks for the answer in boxed notation, I should present it as \boxed{\dfrac{\sqrt{21}}{5}}.

But wait, let me check once more to ensure there's no mistake. The triangle is right-angled at S, so sides are ST = 2, RS = sqrt(21), RT = 5. For angle R, opposite side is ST = 2, hypotenuse RT = 5, so sin R = 2/5. For angle T, opposite side is RS = sqrt(21), hypotenuse RT = 5, so sin T = sqrt(21)/5. All steps check out. Therefore, the answer is correct.
Therefore, $\sin{T} = \boxed{\frac{\sqrt{21}}{5}}$.
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