$\sin R = \dfrac{8}{17}$ means $\dfrac{\text{opposite to }R}{\text{hypotenuse}} = \dfrac{8}{17}$. Decide which sides are “opposite” and “hypotenuse,” then write them as $8k$ and $17k$ for some scale factor $k$.

Once you have two sides written as $8k$ and $17k$, use the Pythagorean theorem to find the third side in terms of $k$. You should get another multiple of $k$.

The area of a right triangle is $\tfrac12 \times (\text{leg}) \times (\text{leg})$. Use your expressions for the legs in terms of $k$, set this equal to $180$, solve for $k$, and then plug into your expression for the hypotenuse.

In Desmos, type sqrt(17^2 - 8^2) to verify that the third side in the $8:17$ right triangle is 15 (so the legs are 8 and 15 in some scale).

Use the fact that the legs are $8k$ and $15k$. In Desmos, enter y = 0.5*8*15*k^2 (which is y = 60k^2) as one expression and y = 180 as another.

Look at the graph of $y = 60k^2$ and the horizontal line $y = 180$. Use the intersection tool or tap the point where they meet to read the value of $k$ that satisfies the area condition.

Once you know $k$, type 17*k into Desmos. The output is the length of the hypotenuse $RS$; match this value to the answer choices.

Angle $T$ is the right angle, so the hypotenuse is the side opposite angle $T$, which is $\overline{RS}$.
For angle $R$:

• The opposite side is $\overline{ST}$.
• The hypotenuse is $\overline{RS}$.

By definition of sine,

So $ST$ and $RS$ must be in the ratio $8:17$.

Let the common scale factor be $k>0$.

• $ST = 8k$
• $RS = 17k$

Now use the Pythagorean theorem to find the third side $RT$:

so $RT = 15k$.
The side lengths are $8k$, $15k$, and $17k$.

In a right triangle, the area is

The legs are $ST = 8k$ and $RT = 15k$, so

We are told the area is $180$ square units, so

Solve for $k$:

We want the length of the hypotenuse $\overline{RS}$, which we wrote as $RS = 17k$. Substitute $k = \sqrt{3}$:

So the hypotenuse $\overline{RS}$ has length \17\sqrt{3}$, which matches choice B.