A vertical pole AB measuring 5 meters snaps at point C. The pole remains in contact at C and the top of the pole touches the ground at point T, a distance of 3 meters from A.

Find the length AC, in meters, the point where the pole snapped.

Guest Dec 27, 2020

#1**0 **

Here we go, my strategy is to solve this as a system of equations, first we label BC, x, and we label AC, y. As given, x+y=5,

using the pythagorean theorem, we find that because $AT$ = 3, then

CA^2+AT^2=CT^2, therefore resulting in y^2+9=x^2, we have two equations:

x+y=5

y^2+9=x^2

can you solve it from here?

Guest Dec 27, 2020

#3**+2 **

Let CA = x

So....CT must just be 5 - x

So....CT is the hypotenuse of a right triangle ....AT is a leg = 3 and CA = x

So.....by the Pyragorean Theorem, we have that

3^2 + x^2 = (5 - x)^2 simplify

9 + x^2 = x^2 - 10x + 25 subtact x^2 from both sides

9 = -10x + 25 rearrange as

25 - 9 = 10x

16 = 10x divide both sides by 10

16/10 = x

8/5 = x = AC = (8/5) m = 1.6 m

CPhill Dec 27, 2020

#4**+3 **

A vertical pole AB measuring 5 meters snaps at point C. The pole remains in contact at C and the top of the pole touches the ground at point T, a distance of 3 meters from A. Find the length AC, in meters, the point where the pole snapped.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 5 AT = 3 AC => **x**

3^{2} + x^{2} = (5 - x)^{2} **x = 1.6 CT = 3.4**

jugoslav Dec 27, 2020