A boat 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)