Chapter 9.8, Problem 27E

To determine

  $To\text{ d}escribe\text{ four methods to find the value of b}\text{.}$

Answer to Problem 27E

  $Thevalue\text{ of }bis21$

Explanation of Solution

Given:

  $\begin{array}{l}\angle B={37}^0\\ AB=35\\ BC=28\end{array}$

Calculation:

  $\begin{array}{l}{1}^{st}\text{ method:}\\ \text{By pythogorous theorem}\\ {\text{(AB)}}^2={\text{(AC)}}^2+{\text{(BC)}}^2\\ \Rightarrow {\text{(35)}}^2={\text{(b)}}^2+{\text{(28)}}^2\\ \Rightarrow {\text{b}}^2={\text{(35)}}^2{\text{-(28)}}^2\\ \Rightarrow {\text{b}}^2=441\\ \Rightarrow \text{b}=21\end{array}$

  $\begin{array}{l}{2}^{nd}\text{ method:}\\ \sin \theta =\frac{opposite\text{ side}}{hypotenuse}\\ \Rightarrow \sin ({37}^0)=\frac{b}{35}\\ \Rightarrow 0.602=\frac{b}{35}\\ \Rightarrow \text{b}=0.602\times 35\\ \Rightarrow \text{b}=21.06\end{array}$

  $\begin{array}{l}{3}^{rd}\text{ method:}\\ \tan \theta =\frac{opposite\text{ side}}{adjacentside}\\ \Rightarrow \tan ({37}^0)=\frac{b}{28}\\ \Rightarrow 0.753=\frac{b}{28}\\ \Rightarrow \text{b}=0.753\times 28\\ \Rightarrow \text{b}=21.09\end{array}$

  $\begin{array}{l}{4}^{th}\text{ method:}\\ \cot \theta =\frac{adjacent\text{ side}}{oppositeside}\\ \Rightarrow \cot ({37}^0)=\frac{28}{b}\\ \Rightarrow 1.327=\frac{28}{b}\\ \Rightarrow \text{b}=\frac{28}{1.327}\\ \Rightarrow \text{b}=21.09\end{array}$