Snapped Pole

$Since\quad \overline { AB } =5\quad meter,\quad Let\quad \overline { AC } =x\quad meter\quad and\quad \\ \overline { CT } =(5-x)meter.\quad Also,\quad \overline { AT } =3\quad meter.\\ Now,\quad in\quad \triangle ACT,\quad By\quad Pythagoras\quad theorem,\\ { AC }^{ 2 }+{ AT }^{ 2 }={ CT }^{ 2 }\quad \Rightarrow \quad { x }^{ 2 }+{ 3 }^{ 2 }={ (5-x) }^{ 2 }\\ \Rightarrow { x }^{ 2 }+9=25+{ x }^{ 2 }-10x\\ \Rightarrow 10x=16\\ \Rightarrow x=1.6$

Cheers!!:):)

We know that BC = TC

Let AC = x ; BC = TC = y

Since AC + BC = AB = 5

y = 5 - x

Now apply pythagoras theorm in ∆ACT

CT^2 = AC^2 + AT^2

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

Solving you get, x = 1.6

let AC=x and given AB=5m then CB=CT=5-x now in right angled triangle CAT , CT^2=AC^2 + AT^2 i.e,( 5-x)^2=x^2 + 3^2 solving we get x=1.6

AC² + AT² = BC²

SINCE BC = CT

BC = 5 – AC

AC² + AT² = (5-AC)²

AC²+AT²= 25- 10AC + AC²

AT= 3

3² = 25 -10AC

9-25 = -10AC

AC = 16/10

AC = 1.6

Let CT = H

And CA. = L

Equations # 1

H . + L = 5

 H^2  - L^2  =. 9

( H . + L) ( H - L ) = 9

( 5 ) ( H - L ) = 9

H - L = 1.8

Equations # 2.

H - L = 1.8

Now we have the system of Equations :

H - L = 1.8

H . + L = 5

2 H = 6.8

H = 3.4

And L = 1.6

Simple.

hypotenuse = (5-x) upper leg = x ground = 3

solve using Pythagorean theorem,

(5-x)^{2} = x^{2} + 3^{2} 25 - 10x + x^2 = x^2 +9 16=10x 1.6=x

AC + CT = 5, If AC= x then CT = 5-x , and AT = 3, ACT is right angled triangle, apply Pythagoras Theorem, CT^2 = AC^2 + AT^2, (5-x)^2 = x^2 + 3^2 , 25 -10x + x^2 = x^2 + 9, 16 = 10x, finally x = 1.6