Chapter 8.5, Problem 17WE

To determine

To find the values of $w$ and $z$

Answer to Problem 17WE

$w=160$ and $z\approx 117$

Explanation of Solution

Given information:

Right angle triangles

It can be observed the triangle contains ${45}^{\circ }$ at one of it vertex with base $=w$ and perpendicular $=160$

Then from tan rule, we have $\tan {45}^{\circ }=\frac{\text{perpendicular}}{\text{base}}$

$\begin{array}{l}\Rightarrow 1=\frac{160}{w}\\ \therefore w=160\end{array}$

The other triangle contains ${30}^{\circ }$ at one of its vertex with same perpendicular $=160$ and base $=w+z,\text{where}w=160$

then from tan rule, we have $\tan {30}^{\circ }=\frac{\text{perpendicular}}{\text{base}}$

  $\begin{array}{l}\Rightarrow \frac{1}{\sqrt{3}}=\frac{160}{w+z}\\ \Rightarrow w+z=160\sqrt{3}\\ \Rightarrow z=160\sqrt{3}-w\\ \Rightarrow z=160\sqrt{3}-160\text{since}w=160\\ \Rightarrow z=160\left(\sqrt{3}-1\right)\\ \Rightarrow z=117.13\\ \therefore z\approx 117\end{array}$

Hence, $w=160$ and $z\approx 117$