Question 436·Hard·Math

0 / 2000

From a point $P$ on level ground, the angle of elevation to the top of a vertical tower is $35^\circ$ . After $P$ is moved $100$ feet directly toward the base of the tower, the angle of elevation becomes $50^\circ$ . Approximately how tall, in feet, is the tower?

Two angles of elevation to one vertical tower create two right triangles with the same height. Use tangent, which compares height with horizontal distance, to write a height equation from each position. Then graph both equations in Desmos in Degrees mode and read their intersection. The distance traveled is the difference between the ground distances, not either full distance.

Hints

Hint 1
The tower and level ground form a right triangle from each viewing point. Tangent compares the vertical height with the horizontal ground distance. If the original distance is $x$ , how can you express the height using the first angle?
Hint 2
Moving toward the base shortens the ground distance by $100$ feet, but the tower's height stays the same. If the original distance is $x$ , what is the distance from the new point?
Hint 3
Each angle gives an equation for the same height. Graph both equations in Degrees mode. At their intersection, the $y$ -coordinate gives the height that fits both viewing points.

Step-by-step

Graph two tangent height equations
Step 1
Write the height from the first angle
Let $x$ be the original ground distance to the tower, and let $y$ be its height. Tangent is the opposite side divided by the adjacent side. From the $35^\circ$ angle, the height is opposite and the ground distance is adjacent, so height equals distance times tangent: $y=x\tan(35^\circ)$ .
Step 2
Use the move to write the second height
Moving $100$ feet toward the base changes the ground distance to $x-100$ ; the tower's height stays $y$ . Apply the same tangent relationship at the new angle: $y=(x-100)\tan(50^\circ)$ . The $100$ feet is the difference between the ground distances, not either full distance.
Step 3
Graph the first height equation
Set Desmos to Degrees because the angles are given in degrees. Type $y=x\tan(35)$ . The line shows the distance-height pairs that fit the first viewing point.
Step 4
Read the height where both equations agree
Add $y=(x-100)\tan(50)$ and click the intersection, where both height equations hold. Desmos shows about $(242.45,169.77)$ . The first coordinate is the original ground distance; the second is the tower's height. So the tower is approximately $170$ feet tall. Choice C.

Question 436·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this