Answer:
[tex] \huge \boxed{ \boxed{ 50 \sqrt{3} \: or \: 86.6}}[/tex]
Step-by-step explanation:
to understand this
you need to know about:
let's solve:
to find the height we need to use tan function because we are given adjacent and angle and we need to figure out opposite (height) [tex]\tan(\theta)=\dfrac{opposite}{adjacent}[/tex]
let opposite be AC
let adjacent be BC
according to the question:
[tex] \quad \: \tan( {60}^{ \circ} ) = \dfrac{AC}{BC} [/tex]
now we need a little bit algebra to figure out AC (height)
- [tex] \sf \: substitute \: the \: given\: value \: of \: BC : \\ \tan( {60}^{ \circ} ) = \frac{AC}{50} [/tex]
- [tex] \sf sustitute \: the \: value \: of \: \tan( {60}^{ \circ} ) \: i.e \: \sqrt{3} : \\ \sqrt{3 } = \frac{AC }{50} [/tex]
- [tex] \sf cross \: multiplication: \\ 50 \sqrt{3} = AC[/tex]
- [tex] \sf swap \: sides \: (your \: wil) : \\ AC = 50 \sqrt{3} \\AC = 86.6 \: ( \sf \: decimal \: if\: needed)[/tex]
[tex]\text{we are done!}[/tex]
