Question

Need help with this trigonometry word problem

A ladder placed against a wall such that it reaches the top of the wall of height 6 m and the ladder is inclined at an angle of 60°. Find how far the ladder is from the foot of the wall.
Please show your work

Answer 1

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.

Answer 2

`\huge \boxed{ \boxed{ \sf 6√(3) \: or \: 10.4}}`

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) = \frac{opp} {adj}`

let adjacent be BC

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

`\quad \: \tan( {60}^( \circ) ) = (6)/(BC)`

we will use a little bit algebra to figure out BC

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

`\text{And we are done!}`