Question

LT1 10th grade level

Answer 1

The distance between the pirate and the treasure can be found from the following relationship between ΔEAF and ΔABC;

`\overline{AE}`/
`\overline{AB}` =
`\overline{FA}`/
`\overline{AC}` =
`\overline{EF}`/
`\overline{BC}`

Explanation:

From the question diagram, we have two triangles, ΔEAF and ΔABC;

The ratio of the lengths of the sides
`\overline{AE}`/
`\overline{AB}` =
`\overline{FA}`/
`\overline{AC}`

∠EAF and ∠BAC are vertical angles, therefore ∠EAF = ∠BAC

Therefore, ΔEAF and ΔABC are similar triangles by Side-Angle-Side, SAS, rule of similarity which states that two triangles that have ratios of a pair of their corresponding sides and the two sides also form equal angles within each triangle, then the two triangles are similar

Therefore, the ratio of the each pair of corresponding sides of the two triangles are equal

We have;

`\overline{AE}`/
`\overline{AB}` =
`\overline{FA}`/
`\overline{AC}` = 50 ft. /(100 ft.) =
`\overline{EF}`/
`\overline{BC}` =
`\overline{EF}`/120 ft.

`\overline{EF}` = 120 ft. × 50 ft./(100 ft.) = 60 ft.

`\overline{EF}` = 60 ft.

The distance between the pirate and the treasure,
`\overline{EF}` = 60 ft.