Chapter 13.4, Problem 42PPS

To determine

To find: Height of the wall.

Answer to Problem 42PPS

Height of the wall $=h=18.9$ feet.

Explanation of Solution

Given information:

Let Point at the base of the wall is B.

Point at the top of the wall be A.

  $\begin{array}{l}B={90}^0\\ S={45}^0\\ L={60}^0\end{array}$

Distance between S and L is $8ft$ .

Given

Let Point at the base of the wall is B.

Point at the top of the wall be A.

  $\begin{array}{l}B={90}^0\\ S={45}^0\\ L={60}^0\end{array}$

Distance between S and L is $8ft$ .

Consider the $\Delta ABL$

Let $x$ be the distance between the base of the wall and Point L Using $\tan \left(\theta \right)=\frac{opposite}{adjacent}$

Using this at Angle L

We get

  $\begin{array}{l}\Rightarrow \tan \left({60}^0\right)=\frac{h}{x}\\ \Rightarrow x=\frac{h}{\tan \left({60}^0\right)}\\ \Rightarrow x=\frac{h}{\sqrt{3}}\end{array}$

Consider the $\Delta ABS$

Let $x$ be the distance between the base of the wall and Point L Using $\tan \left(\theta \right)=\frac{opposite}{adjacent}$

Using this at Angle S

We get

  $\begin{array}{l}\Rightarrow \tan \left({45}^0\right)=\frac{h}{x+8}\\ \Rightarrow 1=\frac{h}{x+8}\\ \Rightarrow x+8=h\\ \Rightarrow x=h-8\end{array}$

So, equating both $x$ values

We get

  $\begin{array}{l}\frac{h}{\sqrt{3}}=h-8\\ \Rightarrow h=\sqrt{3}\left(h-8\right)\\ \Rightarrow h=\sqrt{3}h-8\sqrt{3}\\ \Rightarrow \sqrt{3}h-h=8\sqrt{3}\\ \Rightarrow h\left(\sqrt{3}-1\right)=8\sqrt{3}\\ \Rightarrow h=\frac{8\sqrt{3}}{\sqrt{3}-1}\\ \Rightarrow h=\frac{8\left(1.732\right)}{1.732-1}\\ \Rightarrow h=\frac{13.856}{0.732}\\ \Rightarrow h=18.9\end{array}$

Therefore,

Height of the wall $=h=18.9$ feet.