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Topic
An observer standing at the riverbank wants to measure the width of the river section where it flows through the standing position (the river section is relatively straight, the two banks can be viewed as parallel).
From standing position Ahe measures the angle of inclination \(\alpha \) = 35° relative to the river bank to a position OLD observable on the other side. Then move along the river to the location REMOVE way A some d = 50 m and continue to measure the angle of inclination \(\beta \)=65° relative to the riverbank to the position OLD selected (Figure 53). What is the width of the river that flows through the observer’s position (rounded to the nearest hundredth)?
Solution method – See details
The width of the river is the height drawn from the top OLD byABC
Step 1: Calculate the angle \(\widehat {ABC},\widehat {ACB}\)
Step 2: Use the sine theorem to calculate the length BC byABC
Step 3: Calculate the area of the triangle ABC according to the formula \(S = \frac{1}{2}BC.AB.\sin \widehat {ABC}\)
Step 4: Calculate the height HOLDof the triangle ABC follow the formula \(S = \frac{1}{2}AB.{h_C}\) and then conclude
Detailed explanation
We have: \(\widehat {ABC} = {180^0} – {65^0} = {115^0} \Rightarrow \widehat {ACB} = {180^0} – \left( {\widehat {CAB } + \widehat {ABC}} \right) = {30^0}\)
Apply the sine theorem toABC we have: \(\frac{{BC}}{{\sin A}} = \frac{{AB}}{{\sin C}} \Rightarrow BC = \frac{{AB.\sin A}}{ {\sin C}} = \frac{{50.\sin {{35}^0}}}{{\sin {{30}^0}}} \approx 57.36\) (m)
Triangle area ABC is: \(S = \frac{1}{2}BC.AB.\sin \widehat {ABC} = \frac{1}{2}.57,36.50.\sin {115^0} \approx 1299, 65\) (m2)
Call Hc is the height drawn from the top OLD byABC
We have: \(S = \frac{1}{2}AB.{h_C} \Rightarrow {h_C} = \frac{{2S}}{{AB}} \approx 51.99\) (m)
So the width of the river is 51.99 m
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