A boat heading out to sea starts out at Point A, at a horizontal distance of 639 feet from a lighthouse/the shore. From that point, the boat's crew measures the angle of elevation to the lighthouse's beacon-light from that point to be 11 degrees . At some later time, the crew measures the angle of elevation from point B to be . Find the 3 degrees distance from point A to point B. Round your answer to the nearest foot if necessary .

Respuesta :

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Answer:

  1731 ft

Step-by-step explanation:

The tangent relation is useful in this problem.

  Tan = Opposite/Adjacent

The height of the light is found to be ...

  tan(11°) = height/(639 ft)

  height = (639 ft)·tan(11°) . . . . . solve for height

Then the distance from the light to point B is ...

  tan(3°) = height/(B distance)

  B distance = height/tan(3°) = (639 ft)·tan(11°)/tan(3°) ≈ 2370 ft

Then the distance from point A to point B is ...

  2370 ft -639 ft = 1731 ft . . . AB distance

Tangent or tanθ in a right angle triangle is the ratio of its perpendicular to its base. The distance between points A and B is 1731 feet.

What is Tangent (Tanθ)?

The tangent or tanθ in a right angle triangle is the ratio of its perpendicular to its base. it is given as,

[tex]\rm{Tangent(\theta) = \dfrac{Perpendicular}{Base}[/tex]

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

As it is given that the distance between the lighthouse and point A is 639 feet, while the angle of elevation from that point is 11°. therefore, using the tan function, the height of the lighthouse,

[tex]\rm Tan(\theta) = \dfrac{Perpendicular}{Base}\\\\Tan(11^o) = \dfrac{CD}{CA}\\\\Tan(11^o) = \dfrac{CD}{639}\\\\Tan(11^o) \times 639 = CD\\\\[/tex]

As we know that CD is the height of the lighthouse, now

In Δ BCD

The tan function for the distance between the lighthouse and point B can be written as,

[tex]\rm Tan(\theta) = \dfrac{Perpendicular}{Base}\\\\Tan(3^o) = \dfrac{CD}{CA}\\\\Tan(3^o) = \dfrac{Tan(11^o) \times 639}{CA}\\\\CA = 2370\ ft.[/tex]

Hence, the distance between the lighthouse and Point B is 2370 ft.

Now, we know the distance between the lighthouse and Point A, and the distance between the lighthouse and Point B, therefore,

The Distance between Point A and B (AB) = CB - CA

                                                                       = 2370 - 639

                                                                       = 1731 ft.

Learn more about Tangent (Tanθ):

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