Question 508·Hard·Math

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In right triangle $RST$ , angle $T$ is the right angle. If $\sin R = \dfrac{8}{17}$ and the area of triangle $RST$ is $180$ square units, what is the length of the hypotenuse $\overline{RS}$ ?

A sine ratio compares the leg opposite an angle with the hypotenuse; it doesn't give their actual lengths. If you're also given the area, write the sides using one positive scale factor and put the two legs into the area formula. A restricted Desmos regression can find that factor. Area scales with the square of a side's scale factor, not with the factor itself.

Hints

Hint 1
From $R$ , the opposite leg is $\overline{ST}$ , while $\overline{RS}$ is across from the right angle. Since sine is opposite over hypotenuse, how can you write those two lengths using the same factor $k$ ?
Hint 2
The Pythagorean theorem says the squares of the legs add to the square of the hypotenuse. With sides in an $8$ -to- $17$ ratio, what third number completes the familiar right-triangle pattern?
Hint 3
A right triangle's area is half the product of its perpendicular legs. Both legs contain the same factor $k$ , so their product contains $k^2$ . What positive value of $k$ gives the stated area?

Step-by-step

Scale the triangle, then use its area
Step 1
Identify the sides in the sine ratio
Because $T$ is the right angle, $\overline{RS}$ is the hypotenuse, the side across from it. From $R$ , $\overline{ST}$ is the opposite leg. Sine is opposite over hypotenuse, so:
$\sin R=\frac{ST}{RS}=\frac{8}{17}.$
Step 2
Give both sides one scale factor
The ratio tells you the sides' proportions, not their lengths. Let the positive scale factor be $k$ : $ST=8k,\qquad RS=17k$ . Use the same $k$ for both, so their ratio stays $8$ to $17$ .
Step 3
Find the other leg
The Pythagorean theorem says the legs' squares add to the hypotenuse's square. Since $8^2+15^2=17^2$ , these sides fit the $8$ - $15$ - $17$ pattern. So the other leg is $RT=15k$ .
Step 4
Turn the area into an equation
The legs $ST$ and $RT$ meet at the right angle, so they're the perpendicular legs in the triangle area formula:
$180=\frac12(8k)(15k).$
Multiply the coefficients: $180=60k^2$ . Both legs contain $k$ , so their product contains $k^2$ , not $k$ .
Step 5
Find the positive scale in Desmos
Type $60x_1^2\sim180\{x_1>0\}$ . A regression asks Desmos to find the value of $x_1$ that makes the sides match; here $x_1$ stands for $k$ . The restriction keeps the positive value because lengths can't be negative. Under PARAMETERS, Desmos shows $x_1\approx1.732$ .
Step 6
Keep the scale factor exact
To turn that decimal into an exact length, use the area equation. Divide both sides by $60$ :
$k^2=\frac{180}{60}=3.$
Take the positive square root:
$k=\sqrt3.$
Step 7
Find the requested hypotenuse
Substitute the scale factor into $RS=17k$ : $RS=17\sqrt3$ . The hypotenuse $\overline{RS}$ is $17\sqrt3$ units long. Choice B.

Question 508·Hard·Math

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