Question 580·Hard·Math

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An accessibility ramp forms a right triangle with the ground and the vertical face of a platform. The ramp meets the ground at point $G$ and the platform at point $P$ . The point $C$ is the corner where the ground meets the platform, so $\angle C$ is a right angle and $\overline{GP}$ is the hypotenuse.

A safety cable is attached from $C$ to point $M$ , where $M$ is the midpoint of $\overline{GP}$ . The cable has length 13 feet. The horizontal distance $GC$ is 14 feet greater than the vertical height $CP$ .

Which choice is equal to $\tan G$ ?

In a right triangle, a cable from the right angle to the midpoint of the hypotenuse reaches a point equally far from all three vertices. Double the cable length to get the hypotenuse, then use the difference between the legs in a Pythagorean equation and solve with a Desmos regression. For tangent, divide the leg opposite the named angle by the leg beside it, not by the hypotenuse.

Hints

Hint 1
The midpoint of a right triangle’s hypotenuse is equally far from all three corners. So each half of the hypotenuse is as long as the cable. What is the full ramp length?
Hint 2
Let the platform height be $x$ . The ground length is $14$ more, so it is $x+14$ . The Pythagorean theorem relates the squares of these two legs to the square of the ramp.
Hint 3
Stand at $G$ to name the legs: the vertical leg is opposite that angle, while the ground leg is adjacent. Tangent uses opposite over adjacent, not the slanted ramp.

Step-by-step

Approach 1: Use the midpoint fact and solve in Desmos
Step 1
Find the full ramp length
A midpoint splits a segment into equal halves. In a right triangle, the midpoint of the hypotenuse is also equally far from all three vertices. Since $CM=13$ feet, $GM=MP=13$ feet, so $GP=26$ feet. The cable is half the ramp, not a leg of triangle $GCP$ .
Step 2
Express both legs using the height
Call the vertical height $CP=x$ feet. The ground distance is 14 feet greater, so $GC=x+14$ feet. Because $x$ is a length, $x>0$ .
Step 3
Write the right-triangle equation
The Pythagorean theorem says that the squares of the two legs add to the square of the hypotenuse. The right angle is at $C$ , so $GP$ is the hypotenuse:
$x^2+(x+14)^2=26^2.$
Use $26$ , not $13$ , on the hypotenuse side.
Step 4
Solve for the vertical height
Type $x_1^2+(x_1+14)^2\sim26^2\{x_1>0\}$ in Desmos. The subscript and $\sim$ make this a regression: Desmos finds $x_1$ instead of graphing an equation in $x$ . The restriction keeps the height positive. Under PARAMETERS, Desmos shows $x_1=10$ , so $CP=10$ feet.
Step 5
Find the ground distance
Add $x_1+14$ as the next Desmos line. It displays $24$ , so $GC=24$ feet. That is longer than the height, as the problem says.
Step 6
Form the tangent ratio at G
At $G$ , $CP$ is the opposite leg and $GC$ is the leg beside the angle. Tangent is opposite divided by adjacent, so type $x_1/(x_1+14)$ on the next Desmos line. It shows $0.416666\ldots$ ; its fraction button gives $\frac{5}{12}$ . So $\tan G=\frac{CP}{GC}=\frac{5}{12}$ . Choice A.

Approach 2: Recognize a Pythagorean triple
Step 1
Match the ramp to a known right triangle
Once the midpoint gives $GP=26$ , use the Pythagorean triple $5$ - $12$ - $13$ : doubling its sides gives $10$ - $24$ - $26$ . The legs differ by $14$ , as required. The ground leg is longer, so $CP=10$ and $GC=24$ .
Step 2
Use the triple’s leg ratio
Tangent at $G$ is vertical leg over ground leg. Doubling both legs does not change their ratio, so $\tan G=\frac{10}{24}=\frac{5}{12}$ . Choice A.

Question 580·Hard·Math

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