An accessibility ramp forms a right triangle with the ground and the vertical face of a platform. The ramp meets the ground at point and the platform at point . The point is the corner where the ground meets the platform, so is a right angle and is the hypotenuse. A safety cable is attached from to point , where is the midpoint of . The cable has length 13 feet. The horizontal distance is 14 feet greater than the vertical height . Which choice is equal to ?

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SAT Right Triangles & Trigonometry Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Right Triangles and Trigonometry

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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 $\tan G$ ?

When a right triangle includes the midpoint of the hypotenuse, immediately use the fact that the segment from the right-angle vertex to that midpoint equals half the hypotenuse. That quickly gives the hypotenuse length, after which you can translate any “difference between legs” statement into variables, apply the Pythagorean theorem to solve for the legs, and then form the requested trig ratio.

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Relate $CM$ to the hypotenuse

In a right triangle, the midpoint of the hypotenuse is the same distance from all three vertices. Use that to connect $CM$ and $GP$ .

Use the given difference between legs

Let the vertical height be $h$ . Translate “ $GC$ is 14 greater than $CP$ ” into an equation involving $h$ .

Finish with Pythagorean theorem and tangent

Use $CP^2+GC^2=GP^2$ to solve for the legs, then compute $\tan G=\frac{\text{opposite}}{\text{adjacent}}=\frac{CP}{GC}$ .

Desmos Guide

Find the hypotenuse length

Use the midpoint-of-hypotenuse fact: since $CM=13$ , compute $GP=2\cdot 13$ .

Solve for the leg length

Let $h$ be $CP$ and enter $h^2+(h+14)^2=26^2$ into Desmos. Find the positive solution for $h$ .

Compute the tangent ratio

Compute $GC=h+14$ , then compute $\frac{h}{h+14}$ and match it to the answer choices.

Step-by-step Explanation

Use the midpoint-of-hypotenuse fact

In right triangle $\triangle GCP$ with right angle at $C$ , point $M$ is the midpoint of hypotenuse $\overline{GP}$ . A key property is that the midpoint of the hypotenuse is equidistant from all three vertices, so

CM = \frac{GP}{2}.

Given $CM=13$ , it follows that $GP=26$ .

Set up the legs using the given difference

Let $CP=h$ (the vertical height). Since $GC$ is 14 feet greater than $CP$ , we have

GC = h+14.

Because $GP$ is the hypotenuse,

h^2 + (h+14)^2 = 26^2.

Solve for the height and the horizontal distance

Expand and solve:

h^2 + (h+14)^2 = 676

h^2 + (h^2+28h+196)=676

2h^2+28h-480=0

Divide by 2:

h^2+14h-240=0.

Factor:

(h+24)(h-10)=0.

So $h=10$ (lengths are positive). Then $GC=h+14=24$ .

Compute $\tan G$

Relative to angle $G$ , the opposite side is $CP$ and the adjacent side is $GC$ . Therefore,

\tan G = \frac{CP}{GC} = \frac{10}{24} = \frac{5}{12}.

Therefore, the correct choice is $\frac{5}{12}$ .

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation