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QuestionSam sees Long's Peak at 31∘31^{\circ} from a distance. After moving 1000 ft closer, the angle is 37∘37^{\circ}. Find the height difference.

Studdy Solution

STEP 1

Assumptions1. The initial distance from Sam to the mountain is unknown, we will denote it as dd (in feet). . The angle of elevation from Sam's initial position to the top of Long's Peak is 31∘31^{\circ}.
3. After moving1000 ft closer to the mountain, the new angle of elevation is 37∘37^{\circ}.
4. The height of the mountain above Sam's elevation is unknown, we will denote it as hh (in feet).
5. We assume that the earth is flat in this small region, so we can use simple trigonometry to solve the problem.

STEP 2

We can use the tangent of the angles of elevation to set up two equations. The tangent of an angle in a right triangle is the ratio of the opposite side (the height of the mountain in this case) to the adjacent side (the distance from Sam to the mountain).
The first equation from Sam's initial position istan⁡(31∘)=hd\tan(31^{\circ}) = \frac{h}{d}

STEP 3

The second equation from Sam's position after moving1000 ft closer to the mountain istan⁡(37∘)=hd−1000\tan(37^{\circ}) = \frac{h}{d -1000}

STEP 4

Now we have a system of two equations with two unknowns, hh and dd. We can solve this system by substitution or elimination. Let's use substitution.
First, solve the first equation for hhh=d⋅tan⁡(31∘)h = d \cdot \tan(31^{\circ})

STEP 5

Substitute hh from the first equation into the second equationtan⁡(37∘)=d⋅tan⁡(31∘)d−1000\tan(37^{\circ}) = \frac{d \cdot \tan(31^{\circ})}{d -1000}

STEP 6

implify the equation by multiplying both sides by d−1000d -1000d⋅tan⁡(37∘)=d⋅tan⁡(31∘)−1000⋅tan⁡(31∘)d \cdot \tan(37^{\circ}) = d \cdot \tan(31^{\circ}) -1000 \cdot \tan(31^{\circ})

STEP 7

Rearrange the equation to isolate ddd⋅tan⁡(37∘)−d⋅tan⁡(31∘)=−1000⋅tan⁡(31∘)d \cdot \tan(37^{\circ}) - d \cdot \tan(31^{\circ}) = -1000 \cdot \tan(31^{\circ})

STEP 8

Factor out dd from the left side of the equationd⋅(tan⁡(37∘)−tan⁡(31∘))=−1000⋅tan⁡(31∘)d \cdot (\tan(37^{\circ}) - \tan(31^{\circ})) = -1000 \cdot \tan(31^{\circ})

STEP 9

olve for dd by dividing both sides of the equation by (tan⁡(37∘)−tan⁡(31∘))(\tan(37^{\circ}) - \tan(31^{\circ}))d=−100⋅tan⁡(31∘)tan⁡(37∘)−tan⁡(31∘)d = \frac{-100 \cdot \tan(31^{\circ})}{\tan(37^{\circ}) - \tan(31^{\circ})}

STEP 10

Now that we have dd, we can substitute it back into the first equation to find hhh=d⋅tan⁡(31∘)h = d \cdot \tan(31^{\circ})

STEP 11

Substitute the value of dd from step9 into the equationh=−1000⋅tan⁡(31∘)tan⁡(37∘)−tan⁡(31∘)⋅tan⁡(31∘)h = \frac{-1000 \cdot \tan(31^{\circ})}{\tan(37^{\circ}) - \tan(31^{\circ})} \cdot \tan(31^{\circ})

STEP 12

implify the equation to find the value of hh. This will give us the height of the top of Long's Peak above Sam's elevation.
The top of Long's Peak is higher than Sam's elevation by hh feet.

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