Trigonometry:

The apparent size of a tall building 2 km away is 0.05 radians. What is the building's approximate height?



the answer was 100 m but I was just wondering how they got 100​

TrigonometryThe apparent size of a tall building 2 km away is 005 radians What is the buildings approximate heightthe answer was 100 m but I was just wondering class=

Respuesta :

A. The approximate height of the building is 100 m

B. We can use the arc length formula to obtain an approximation of BC since angle A is small.

A.

The approximate height of the building is 100 m

To find the approximate height of the building, we consider the diagram.

From the diagram, we have that using trigonometry,

tanA = BC/AB

Now

  • A = 0.05 radians,
  • BC = height of building and
  • AB = 2 km = 2000 m

Finding the value of BC

So, making BC subject of the formua, we have

BC = ABtanA

Substituting the values of the variables into the equation, we have

BC = ABtanA

BC =2000tan0.05

BC = 2000 × 0.05

BC = 100 m

So, the approximate height of the building is 100 m

B.

We can use the arc length formula to obtain an approximation of BC since angle A is small.

Arc Length Formula

We know that the arc length formula L = rФ where

  • r = radius and
  • Ф = angle in radians

The approximate height

Now BC = ABtanA

We know that Ф ≅ tanФ when Ф is small.

The comparison

So, BC = AB × A which is the arc length formula with

  • L = BC,
  • r = AB and
  • Ф = A

So, we can use the arc length formula to obtain an approximation of BC since angle A is small.

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