The correct answer is A. 27.8 ft
We can use trigonometry to solve this problem.
At low tide:
tan(angle) = opposite/adjacent
tan(angle) = 20/40
angle = arctan(0.5) = 26.57 degrees
At high tide:
Now, the adjacent side is 35 ft and we know the angle is 26.57 degrees. We can solve for the opposite side (height of boat above seafloor) using tan(angle) = opposite/adjacent
tan(26.57) = opposite/35
opposite = 35 * tan(26.57)
opposite = 27.8 ft
Therefore, the boat is floating approximately 27.8 ft above the seafloor at high tide.
Select the correct answer.
A boat is anchored in the water. The anchor lies at point A. At low tide, the boat is floating 20 feet above the seafloor and is a horizontal distance of 40 feet away from its anchor. At high tide, the boat is a horizontal distance of 35 feet away from its anchor.
An image shows two right-angled triangles, named low tide and high tide. The length of the opposite and adjacent sides of low tide triangles is 20 ft. and 40 ft. The length of the adjacent side of the high tide triangle is 35 ft.
Approximately how high is the boat floating above the seafloor at high tide? Assume the anchor rope remains tight, without any slack.
A. 27.8 ft
B. 35.7 ft
C. 22.5 ft
D. 15.6 ft
1 answer
1 answer