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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 *A*he 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* by*ABC*

Step 1: Calculate the angle \(\widehat {ABC},\widehat {ACB}\)

Step 2: Use the sine theorem to calculate the length *BC* by*ABC*

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 *H _{OLD}*of 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 to*ABC *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: \(S = \frac{1}{2}BC.AB.\sin \widehat {ABC} = \frac{1}{2}.57,36.50.\sin {115^0} \approx 1299, 65\) (m^{2})

Call *H _{c}* is the height drawn from the top

*OLD*by

*ABC*

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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