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.
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?
+128407 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
+1639 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.
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AB = 5 AT = 3 AC => x
32 + x2 = (5 - x)2 x = 1.6 CT = 3.4
