Question

Points A and B are separated by a lake. To find the distance between them, a surveyor locates a point Con land such that .CAB = 48.4º. He also measures CA as 313 ft and CB as 527 ft. Find the distance between A and B. (Round your answer to one decimal place.) ft

Answer 1

Since we have three points: A, B and C we can locate them in the vertexes of a triangle.

Now, from the observations made by the surveyor we know that the angle Notice that we also name the angles and distances. To find c, that is the distance between A and B, we can use the Law of sines

`(\sin\alpha)/(a)=(\sin\beta)/(b)=(\sin \gamma)/(c)`

but first we have to find the angle alpha. From the first two terms of the law we know that

`(\sin\alpha)/(a)=(\sin \beta)/(b)`

Plugging the values we have

`\frac{\sin\alpha}{313\text{ ft}}=\frac{\sin 48.4}{527\text{ ft}}`

Solving for alpha, we have

`\begin{gathered} \frac{\sin\alpha}{313\text{ ft}}=\frac{\sin 48.4}{527\text{ ft}} \\ \sin \alpha=\frac{313\text{ ft}}{527\text{ ft}}\sin 48.4 \\ \alpha=\sin ^(-1)((313)/(527)\sin 48.4) \\ \alpha=26.4^(\circ) \end{gathered}`

Once we have the angle alpha, we can calculate the angle gamma. We know that the sum of all the interior angles of a triangle have to be 180, then

`\alpha+\beta+\gamma=180`

Plugging the values we know and solving for gamma we have

`\begin{gathered} \alpha+\beta+\gamma=180 \\ 26.4+48.4+\gamma=180 \\ \gamma=180-26.4-48.4 \\ \gamma=105.2 \end{gathered}`

Finally we can calculate the distance between the points A and B using the law of sines once again.

`(\sin\beta)/(b)=(\sin \gamma)/(c)`

Substituing the values we know and solving for c, we have

`\begin{gathered} \frac{\sin48.4}{527\text{ ft}}=(\sin105.2)/(c) \\ c\frac{\sin48.4}{527\text{ ft}}=\sin 105.2 \\ c=(\sin105.2)/(\sin48.4)(527\text{ ft)} \\ c=680.1\text{ ft} \end{gathered}`

Therefore the distance between the points A and B is 680.1 ft.