Answer:
[tex] \huge \boxed{ \boxed{ \sf 6\sqrt{3} \: or \: 10.4}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
let's solve:
to find how far the ladder is from the foot of the wall
we will use tan function because we are given opposite and angle
we need to find adjacent
[tex] \quad \tan( \theta) = \dfrac{opp} {adj} [/tex]
let adjacent be BC
[tex] \boxed{ \red{ \boxed{accoding \: to \: the \: question : }}}[/tex]
[tex] \quad \: \tan( {60}^{ \circ} ) = \dfrac{6}{BC} [/tex]
we will use a little bit algebra to figure out BC
- [tex] \sf substitute \: the \: value \: of \: \tan( {60}^{ \circ} ) \: i.e \: \sqrt{3} : \\ \sqrt{3} = \frac{BC}{6} [/tex]
- [tex] \sf cross \: multiplication : \\ \therefore \: BC = 6\sqrt{3} \: or \: 10.4[/tex]
[tex]\text{And we are done!}[/tex]
