Chapter 14.4, Problem 32PPSE

Solution

To find: The perimeter of a lot having shape of an isosceles triangle.

The perimeter of the lot is $\text{504}\text{.4}\text{}\text{ft}$

Given data: The vacant lot is of the form of an isosceles triangle between two streets intersecting at angle ${85.9}^{\text{ο}}$ the distance between ends of these streets is 150 ft long.

Method/Formula used:

An isosceles triangle ABC is shown in Fig. 1. If $\angle BAC=\theta$ , $\text{AB}=c,\text{BC}=a,CA=b$ , then $\cos \theta =\frac{a^2+b^2-c^2}{2ab}$ … (1)

  

Calculations:

Two streets AB and AC each of 150 ft long are two equal sides of lot that is in the form of isosceles triangle ABC shown in Fig.2. The sides AB and AC intersects at angle ${85.9}^{\text{ο}}$ .

  

From Fig. 2, in triangle ABC, from result (1),

  $\cos \angle \text{A}=\frac{b^2+c^2-a^2}{2bc}$

Substitute ${85.9}^{\text{ο}}$ for $\angle \text{A}$ , 150 for each b and c ,

  $\cos \left({85.9}^{\text{ο}}\right)=\frac{{150}^2+{150}^2-a^2}{2\left(150\right)\left(150\right)}$

  $\begin{array}{c}\cos \left({85.9}^{\text{ο}}\right)=\frac{22500+22500-a^2}{2\cdot \left(22500\right)}\\ =\frac{45000-a^2}{45000}\end{array}$

Substitute 0.0715 for $\cos {85.9}^{\text{ο}}$ ,

  $\begin{array}{c}\left(0.0715\right)\cdot 45000=45000-a^2\\ 3217.5=45000-a^2\\ a=\sqrt{45000-3217.5}\\ a=204.41\end{array}$

Thus, the side BC of the lot is $204.4$ ft. Therefore, the perimeter of the lot is,

  $\begin{array}{c}\text{Perimeter=150}\text{ft+150 ft+204}\text{.4 ft}\\ \text{=504}\text{.4}\text{}\text{ft}\end{array}$