Trigonometry - SOHCAHTOA - Pythagoras

Question

A yacht is moored to the side of a quay by a rope 3 metres long. The rope is tied to the yacht at R and the quay at T.

At low tide, when the yacht is as far away from the quay as possible, the rope makes an angle of 64 degrees with the horizontal.

Note : Line R to T = 3 metres. This is the hypotenuse as it is opposite a 90 degree angle.

Calculate the vertical distance of R below T when the yacht is in this position. Give answer correct to one decimal place.

At high tide, the water level at the quay is 1.6 metres higher than at low tide.

At high tide, when the yacht is as far away from the quay as possible, calculate the distance from R to the side of the quay.

Answer 1

The hypotenuse is always 3 m. Since the angle the rope makes with the horizontal is given, we can use the trig ratios to find the leg lengths of the right triangle formed from the rope, the vertical distance below the mooring point, and the horizontal distance from the quay.

Let V equal the vertical distance of the boat (below its mooring point), H be its horizontal distance from the quay, and L be the length of the rope.

Then sin(64 deg) = V/L and cos(64 deg) = H/L ==> V=L*sin(64) and H=L*cos(64).

For part b, subtract 1.6 from the value of V found in part A and use the Pythagorean theorem to find H.

Answer 2

Draw a figure!

At low tide: distance = 3 sin 64 meters)

At least if I understand the 2nd part of the question, the answer should still be 3...but maybe I am misunderstanding the question.