Aboat drops an anchor 75 feet to the bottom of a lake. a sunken treasure chest lies at the bottom of the lake, 40 feet away from the anchor. what is the distance from the boat to the treasure chest ?

Given
Height (length) of the anchor dropped from the boat = 75 feet
Distance of the treasure from the anchor = 40 feet

Solution
The dropped at the bottom makes an angle of 90 degree . Therefore making an right triangle with the chest and boat

By Pythagoras theorem
AC^2 = AB^2 + BC^2

(AC = distance of the treasure from the boat
AB = height of the anchor dropped from the boat
BC = the distance of treasure from the anchor)

AC^2 = 75^2 + 40^2
AC^2 = 5325 +1600
AC^2 = 6925
AC = 25 * (under root 11)

Therefore they have cover a distance of 25 * (under root 11)

Given
Height (length) of the anchor dropped from the boat = 75 feet
Distance of the treasure from the anchor = 40 feet

Solution
The dropped at the bottom makes an angle of 90 degree . Therefore making an right triangle with the chest and boat

By Pythagoras theorem
AC^2 = AB^2 + BC^2

(AC = distance of the treasure from the boat
AB = height of the anchor dropped from the boat
BC = the distance of treasure from the anchor)

AC^2 = 75^2 + 40^2
AC^2 = 5325 +1600
AC^2 = 6925
AC = 25 * (under root 11)

Therefore they have cover a distance of 25 * (under root 11)