Question

Estimating the Width of the Mississippi River A tourist at the top of the Gateway Arch (height, 630 feet) in St. Louis, Missouri, observes a boat moored on the Illinois side of the Mississippi River 2070 feet directly across from the Arch. She also observes a boat moored on the Missouri side directly across from the first boat (see diagram). Given that $B={\cot }^{-1}\left(\frac{67}{55}\right)$, estimate the width of the Mississippi River at the St. Louis riverfront.

Short answer

The length of the river is 1650 ft.

Step-by-step solution

Step 1. Given information

The information given in the question is plotted in a diagram follows:

We are given with the length DI = 2070 ft. , the height of the gateway arch i.e., CD = 630 ft. and$\cot \left(B\right)=\frac{67}{55}.$

Step. 2 Analyzing the question

In the question we are asked to find the length of the river which is MI in our diagram. So to find MI we have following equations:

$$\begin{gathered} In∆DCIwehave, \\ \tan \left(A+B\right)=\frac{DI}{CD}=\frac{2070}{630}=\frac{23}{7}.-----\left(1\right) \\ Now,wehave\cot \left(B\right)=\frac{67}{55}, \\ So,\tan \left(B\right)=\frac{55}{67}.-----\left(2\right) \\ AndIn∆DCM, \\ \tan \left(A\right)=\frac{DM}{CD}=\frac{DI-MI}{630}.-----\left(3\right). \end{gathered}$$

Step 3. Find tanA.

As we know $\tan \left(A+B\right)=\frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}.$

Using this equation (1),

$$\begin{gathered} \frac{\tan \left(A\right)+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}=\frac{23}{7}, \\ 7\tan \left(A\right)+7\tan \left(B\right)=23-23\tan \left(A\right)\tan \left(B\right), \\ \tan \left(A\right)=\frac{23-7\tan \left(B\right)}{7+23\tan \left(B\right)}, \\ Nowfromequation\left(2\right)weknowthevalueof\tan \left(B\right). \\ So,puttingthisvalueinaboveweget, \\ \tan \left(A\right)=\frac{23-7\times {\displaystyle \frac{55}{67}}}{7+23\times {\displaystyle \frac{55}{67}}}=\frac{1541-385}{469+1265}=\frac{1156}{1734}. \end{gathered}$$

Step 4. Finding the length of the river i.e., MI

$$\begin{gathered} Nowwehave, \\ \tan \left(A\right)=\frac{1156}{1734}, \\ and\tan \left(A\right)=\frac{DI-MI}{630}.(Fromequation(3\left)\right) \\ Now,fromabovewecanwrite, \\ \frac{DI-MI}{630}=\frac{1156}{1734}, \\ FromdiagramweknowDI=2070ft. \\ SosolvingforMIweget, \\ 2070-MI=\frac{1156\times 630}{1734}, \\ MI=2070-\frac{1156\times 630}{1734}, \\ MI=2070-420=1650ft. \end{gathered}$$

Therefore, The length of the river will be 1650 ft.