Question

Need help with this trigonometry word problem

Answer 1

Answer:

$\huge \boxed{ \boxed{ \sf 6\sqrt{3} \: or \: 10.4}}$

Step-by-step explanation:

to understand this

you need to know about:

• trigonometry
• PEMDAS

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

$\quad \tan( \theta) = \dfrac{opp} {adj}$

let adjacent be BC

$\boxed{ \red{ \boxed{accoding \: to \: the \: question : }}}$

$\quad \: \tan( {60}^{ \circ} ) = \dfrac{6}{BC}$

we will use a little bit algebra to figure out BC

1. $\sf substitute \: the \: value \: of \: \tan( {60}^{ \circ} ) \: i.e \: \sqrt{3} : \\ \sqrt{3} = \frac{BC}{6}$
2. $\sf cross \: multiplication : \\ \therefore \: BC = 6\sqrt{3} \: or \: 10.4$

$\text{And we are done!}$

Answer 2

Step-by-step explanation:

in triangle ABC

relationship between perpendicular and base is given by tan angle

tan 60=p/b

b=6/tan60=3.46m

the ladder is 3.46m from the foot of the wall.