contestada

Captain Sparrow leaves Pirates Cove along a bearing of S 30° E. He travels at a rate of
30 miles per hour for 3 hours and then drops anchor.
A) Determine the distance east and the distance south from Pirates Cove of Captain
Sparrow's ship when he drops anchor. Be sure to show the equations you used to find
your answers. Round to three decimal places.
B) Shipwreck Island is located 80 miles east and 20 miles south of Pirates Cove.
Determine the distance from Captain Sparrow's ship to Shipwreck Island. Be sure to
justify your answer. Round to 3 decimal places

Respuesta :

A) The distance east from Pirates Cove of Captain Sparrow's ship when he drops the anchor is 45 miles

The distance south from Pirates Cove of Captain Sparrow's ship when he drops the anchor is 77.942 miles

B) The distance from Captain Sparrow's ship to Shipwreck Island is 67.693 miles

Trigonometry

In the illustrative diagram,

C represents the Pirates Cove

S represents Captain Sparrow's ship when he drops the anchor

and I represent Shipwreck Island

N, E, W, and S represent the four cardinal points

First, we will determine the distance Captain Sparrow travels

Using the formula,

Distance = Speed × Time

∴ Distance = 30 miles/hour × 3 hours
Distance = 90 miles

This the distance Captain Sparrow covers when he drops the anchor

A)

The distance east from Pirates Cove of Captain Sparrow's ship when he drops the anchor is given by x

Using SOH CAH TOA

[tex]sin 30^\circ = \frac{x}{90}[/tex]

y = 90 × sin 30°

y = 45 miles

∴ The distance east from Pirates Cove of Captain Sparrow's ship when he drops the anchor is 45 miles

The distance south from Pirates Cove of Captain Sparrow's ship when he drops the anchor is given by y

Using SOH CAH TOA

[tex]cos 30^\circ = \frac{y}{90}[/tex]

y = 90 × cos 30°

y = 77.942 miles

∴ The distance south from Pirates Cove of Captain Sparrow's ship when he drops the anchor is 77.942 miles

B)

The distance from Captain Sparrow's ship to Shipwreck Island is given by /SI/

First, we will determine /CI/

By the Pythagorean theorem,

/CI/² = 80² + 20²

/CI/² = 6400 + 400

/CI/² = 6800

/CI/ = √6800

/CI/ = 82.4621 miles

Now, we will determine ∠ICA

Let angle ICA = θ

[tex]tan \theta = \frac{20}{80}[/tex]

tanθ = 0.25

θ = tan⁻¹ (0.25)

θ = 14.036°

∴ ∠SCI = 90° - (30° + 14.036°)

∠SCI = 90° - 44.036°

∠SCI = 45.964°

Now, consider Δ SCI

Using the laws of cosine

/SI/² = /SC/² + /CI/² - 2×/SC/×/CI/ cos(SCI)

/SI/² = 90² + 6800 - 2 × 90 × 82.4621 cos45.964°

/SI/² = 8100 + 6800 - 14843.178 cos45.964°

/SI/² = 14900 - 14843.178 cos45.964°

/SI/² = 14900 - 10317.6445

/SI/² = 4582.3555

/SI/ = √4582.3555

/SI/ = 67.693 miles

Hence, the distance from Captain Sparrow's ship to Shipwreck Island is 67.693 miles

Learn more on Trigonometry and Bearings & Distances here: https://brainly.com/question/22518031

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