Question

A ladder is placed against a wall such that it touches it, 20 m above the ground. The foot of the
ladder is 15 m away from the wall. What is the length of the ladder?

Answer 1

Answer:

z=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5iz=−16+30i=−16+2(15i)=9−25+2(3)(5i)=3

2

+(5i)

2

+2(3)(5i)=(3+5i)

2

z

=3+5i or −3−5i