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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.

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?

Dec 27, 2020
#2
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Ok no need, the answers from my work show that

x=18/5

y=7/5

Dec 27, 2020
#3
+117724
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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

Dec 27, 2020
edited by CPhill  Dec 27, 2020
#4
+1164
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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

32 + x2 = (5 - x)2          x = 1.6           CT = 3.4

Dec 27, 2020