Approximately what is the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertical height of 150 m, as shown in the diagram opposite?

The set up will be equivalent to a right angled triangle where the height is the opposite side facing the 45° angle directly. The length of the rope will be the slant side which is the hypotenuse.

Using the SOH, CAH, TOA trigonometry identity to solve for the length of the rope;

Since we have the angle theta = 45° and opposite = 150m

According to SOH;

Sin theta = opposite/hypotenuse.

Sin45° = 150/hyp

hyp = 150/sin45°

hyp = 150/(1/√2)

hyp = 150×√2

hyp = 150√2 m

hyp = 212.13m

Hence the length of the rope for the kite sail, in order to pull the ship at an angle of 45° and be at a vertical height of 150 m is approximately 212m

akposevictor akposevictor · Answer 2 · 2021-11-09T18:44:26+01:00

By applying trigonometry ratio, the length of the rope for the kite to sail would be: 212 m.

Recall:

Trigonometry ratios used to solve a right triangle are: SOH CAH TOA

The diagram describing the situation is attached below (see attachment).

Thus:

The reference angle [tex](\theta) = 45^{\circ}[/tex]

Let the length of the rope be x = Hypotenuse

Opposite = 150 m

To find the length of the rope (x), apply SOH

Thus:

[tex]Sin(\theta) = \frac{Opp}{Hyp}[/tex]

Substitute

[tex]Sin(45) = \frac{150}{x}[/tex]

Multiply both sides by x

[tex]x \times Sin(45) = \frac{150}{x} \times x\\\\x \times Sin(45) = 150[/tex]

Divide both sides by sin(45)

[tex]\frac{x \times Sin(45)}{Sin(45)} = \frac{150}{ Sin(45)} \\\\\mathbf{x = 212 $ m}[/tex]

Therefore, by applying trigonometry ratio, the length of the rope for the kite to sail would be: 212 m.

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