Question

to find the distance across a river, a surveyor choose points a and b, which are 238 m apart on one side of the river. she then chooses a reference point c on the opposite side of the river and finds that ∠bac ≈ 83° and ∠abc ≈ 50°. note: the picture is not drawn to scale. approximate the distance from point a to point c. distance = \boxed{} m find the distance across the river. height = \boxed{} m enter your answer as a number; your answer should be accurate to 2 decimal places. question help: message instructor

Explanation:

Step1: Find angle \(C\)

In \(\triangle ABC\), we know that the sum of angles in a triangle is \(180^{\circ}\). So, \(\angle C=180^{\circ}-\angle BAC - \angle ABC\).
Substituting \(\angle BAC = 83^{\circ}\) and \(\angle ABC=50^{\circ}\), we get \(\angle C=180^{\circ}-83^{\circ}-50^{\circ}=47^{\circ}\)

Step2: Use the Law of Sines to find \(AC\)

The Law of Sines states that \(\frac{AB}{\sin C}=\frac{AC}{\sin B}\)
We know \(AB = 238\space m\), \(\angle B = 50^{\circ}\), \(\angle C=47^{\circ}\)
So, \(AC=\frac{AB\times\sin B}{\sin C}\)
Substitute the values: \(AC=\frac{238\times\sin(50^{\circ})}{\sin(47^{\circ})}\)
Since \(\sin(50^{\circ})\approx0.7660\) and \(\sin(47^{\circ})\approx0.7314\)
\(AC=\frac{238\times0.7660}{0.7314}=\frac{182.208}{0.7314}\approx249.12\space m\)

Step3: Find the distance across the river (height \(h\) from \(C\) to \(AB\))

Let the height be \(h\). We know that \(\sin A=\frac{h}{AC}\) (using right - triangle trigonometry, if we drop a perpendicular from \(C\) to \(AB\))
Since \(AC\approx249.12\space m\) and \(\angle A = 83^{\circ}\), \(\sin(83^{\circ})\approx0.9925\)
\(h = AC\times\sin A\)
\(h=249.12\times\sin(83^{\circ})\)
\(h = 249.12\times0.9925\approx247.25\space m\)

Answer:

distance \(= 249.12\space m\)
height \(= 247.25\space m\)