Question

A 26-inch skateboard leans against a wall making a 39angle with the ground

Answer 1

Diagram:-

`\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf \!\!\!\!Wall}\put(2.8,.3){\large\bf Ground}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf$39^(\circ)$ }\end{picture}`

We need the walls height then ground distance

`\\ \rm\Rrightarrow sin\theta=(Perpendicular)/(Hypotenuse)`

`\\ \rm\Rrightarrow sin39=(P)/(26)`

`\\ \rm\Rrightarrow P=26sin39`

`\\ \rm\Rrightarrow P=16.4in`

Distance of wall from skateboard touching point on ground:-

`\\ \rm\Rrightarrow cos\theta=(Base)/(Hypotenuse)`

`\\ \rm\Rrightarrow cos39=(B)/(26)`

`\\ \rm\Rrightarrow B=26cos39`

`\\ \rm\Rrightarrow B=20.2in`

#BBH