A building is 53 feet high. At a distance away from the building, a 6 foot tall observer notices the angle of elevation to the top of the building is 43°. How far is the observer from the base of the building?

thankyou x

If a person stands at a point and looks up at an object, the angle between their horizontal line of sight and the object is called the angle of elevation.

To find how far the observer is from the base of the building, model as a right triangle and solve using the tan trigonometric ratio.

Tan trigonometric ratio

[tex]\sf \tan(\theta)=\dfrac{O}{A}[/tex]

where:

[tex]\theta[/tex] is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given information:

Building height = 53 ft
Person height = 6 ft
Angle of elevation = 43°

As the person is 6 ft tall, we model the line of sight as 6 ft above ground level. Therefore, the side of the right triangle opposite the angle of elevation is:

⇒ 53 ft - 6 ft = 47 ft

Substituting the found/given values into the formula and solving for A:

[tex]\implies \sf \tan(43^{\circ})=\dfrac{47}{A}[/tex]

[tex]\implies \sf A=\dfrac{47}{\tan(43^{\circ})}[/tex]

[tex]\implies \sf A=50.40132937...[/tex]

Therefore the observer is 50.4 ft (nearest tenth) from the base of the building.

A building is 53 feet high. At a distance away from the building, a 6 foot tall observer notices the angle of elevation to the top of the building is 43°. How far is the observer from the base of the building? ​thankyou x

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A building is 53 feet high. At a distance away from the building, a 6 foot tall observer notices the angle of elevation to the top of the building is 43°. How far is the observer from the base of the building?

thankyou x