SOLUTION:
Step 1:
In this question, we are given the following:
Step 2:
The details of the solution are as follows:
The width of the river can see calculated thus:
[tex]\begin{gathered} \\ Using\text{ Trignometry, we have that:} \\ tan\text{ 29.2}^0\text{ =}\frac{opposite\text{ }}{adjacent}=\frac{y}{116} \end{gathered}[/tex][tex]\begin{gathered} cross-multiply,\text{ we have that:} \\ y\text{ = 116 x tan 29.2}^0 \\ Then,\text{ } \\ y\text{ = 116 x 0.5589} \\ y\text{ = 64.8324 m} \\ y\text{ }\approx\text{ 65 m \lparen to the nearest metre\rparen} \end{gathered}[/tex]CONCLUSION:
The width of the river is:
[tex]y=\text{ 65 m \lparen correct to the nearest metre\rparen}[/tex]