A ladder $20 m$ long touches a window of a tower which is at height of $15 m$ from earth surface then find the distance between foot of the tower and ladder:

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Solution

Let $A C$ be the ladder
$A B$ be distance between foot of tower and ladder
using Pythagoras theorem in right triangle
${A C}^{2} = {A B}^{2} + B C^{2}$
$20^{2} = {A B}^{2} + 15^{2}$
$20^{2} - 15^{2} = {A B}^{2}$
$175 = {A B}^{2}$
$A B = 5 \sqrt{7}$ m

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A ladder 20 m long touches a window of a tower which is at height of 15 m from earth surface then find the distance between foot of the tower and ladder:

Let AC be the ladderAB be distance between foot of tower and ladderusing Pythagoras theorem in right triangle AC2=AB2+BC2202=AB2+152202−152=AB2175=AB2AB=5√7m

A ladder $20 m$ long touches a window of a tower which is at height of $15 m$ from earth surface then find the distance between foot of the tower and ladder:

Let $A C$ be the ladder
$A B$ be distance between foot of tower and ladder
using Pythagoras theorem in right triangle
${A C}^{2} = {A B}^{2} + B C^{2}$
$20^{2} = {A B}^{2} + 15^{2}$
$20^{2} - 15^{2} = {A B}^{2}$
$175 = {A B}^{2}$
$A B = 5 \sqrt{7}$ m