Question

A tower on a college campus was built with a faulty foundation and is starting to lean. A student climbs to the tilted top and drops a rope down to the ground. The
end of the rope lands 3 feet from the base of the tower and measures 57 feet from
the top of the building to the ground. What is the angle that the tower is leaning?
Round your answer to the nearest degree, and enter the number only

Answer 1

Answer:

leaning at 3 degrees off of normal (to the nearest whole degree)

Step-by-step explanation:

To visualize, let's assume the tower is leaning to the right, (as viewed by us).

If the rope is dropped from the top to the ground, the tower, the rope and the ground (between the rope and the tower) form a right triangle.

The height and base of the triangle are given

Often we're looking for the angle near the ground, but here, we're looking for the angle between the rope and the tower because the angle is congruent to the angle that the tower forms with it's original vertical position (alternate interior angles).  If the tower was standing straight up and hadn't been leaning at all, we wouldn't say that the tower was leaning 90°... we would say it was leaning 0°.  Whereas if the tower fell over and was laying flat on the ground, we would say it was leaning at 90°.  So, we're not measuring the angle between the tower and the ground, but rather the angle between the tower where it should be, and where it is... which is the same angle as the rope forms with the tower.

Recall that $tan(\theta)=\dfrac {opp}{hyp}$

Since we know the two legs of the right triangle, one can setup a tangent relationship with the two legs, but remember that the "opposite" side is going to be the ground, and the "adjacent side" will be the rope.

$tan(\theta)=\dfrac {ground}{rope}\\tan(\theta)=\dfrac {3ft}{57ft}\\tan(\theta)=\frac {3}{57}\\\theta=tan^{-1}(\frac {3}{57})\\\theta=3.012787504^o\\\theta \approx 3^o$