Question

What is the measure of AB? (TO THE NEAREST KILOMETER)

Answer 1

Let's make a skecth of the triangle in the figure:

We can look at the two triangles separatelly. The only thing the they have in common is the angle ∠S, so we can use this as our information to extract from one and use in the other.

We know all sides of triangle SCD, so we can use the Law of Cosines to find out the cosine of angle ∠S:

`\begin{gathered} CD^2=SC^2+SD^2-2\cdot SC\cdot SD\cos \angle S \\ 2\cdot SC\cdot SD\cos \angle S=SC^2+SD^2-CD^2 \\ \cos \angle S=(SC^2+SD^2-CD^2)/(2\cdot SC\cdot SD) \end{gathered}`

So, inputting the values, we have:

`\begin{gathered} \cos \angle S=(586^2+644^2-533^2)/(2\cdot586\cdot644) \\ \cos \angle S=(343396+414736-284089)/(754768) \\ \cos \angle S=(343396+414736-284089)/(754768)=\frac{474043}{754768{}} \\ \cos \angle S=\frac{474043}{754768{}} \\ \cos \angle S=0.628064\ldots \end{gathered}`

Now, we can look to the greater triangle. Since this angle is the same, we can use the Law of Cosines again, but now to find out the length of AB:

`AB^2=SA^2+SB^2-2\cdot SA\cdot SB\cdot\cos \angle S`

The measures are:

`\begin{gathered} SA=SC+CA \\ SA=586+733 \\ SA=1319 \\ SB=SD+DB \\ SB=644+800 \\ SB=1444 \end{gathered}`

So:

`\begin{gathered} AB^2=SA^2+SB^2-2\cdot SA\cdot SB\cdot\cos \angle S \\ AB^2=1319^2+1444^2-2\cdot1319\cdot1444\cdot0.628064\ldots \\ AB^2=1739761+2085136-2392468.58\ldots \\ AB^2=1432428.41\ldots \\ AB=\sqrt[]{1432428.41\ldots} \\ AB=1196.84\ldots \\ AB\approx1197 \end{gathered}`

So, the measure of AB is approximately 1197 m, which is 1.197 km, so the measure of AB to the nearest kilometer is approximately ilo1 km.