Answer:
[tex]63\ \text{ft}[/tex]
Step-by-step explanation:
The distances are shown in the figure
[tex]\tan25^{\circ}=\dfrac{p}{b}\\\Rightarrow p=b\tan25^{\circ}[/tex]
[tex]\tan15^{\circ}=\dfrac{p}{b+100}\\\Rightarrow p=b\tan15^{\circ}+100\tan15^{\circ}\\\Rightarrow b\tan25^{\circ}=b\tan15^{\circ}+100\tan15^{\circ}\\\Rightarrow b=\dfrac{100\tan15^{\circ}}{\tan25^{\circ}-\tan15^{\circ}}\\\Rightarrow b=135.08\ \text{ft}[/tex]
[tex]p=b\tan25^{\circ}=135.08\times\tan25^{\circ}\\\Rightarrow p=62.99\approx 63\ \text{ft}[/tex]
The width of the river is [tex]63\ \text{ft}[/tex].