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g100num [7]
2 years ago
5

A blimp is 1600 meters high in the air and measures the angles of depression to two stadiums to the east of the blimp. if those

measurements are 71.7° and 25.2°, how far apart are the two stadiums?
Mathematics
1 answer:
Sonja [21]2 years ago
5 0

The two stadiums are 2,871 meters

A blimp    = point A

stadium 1  = point B

stadium 2 = point C

height of the blimp ,  AD = 1600

depression angle of stadium 1 , ∠y = 71.7°

depression angle of stadium 2 , ∠x = 25.2°

distance between the two stadiums  = d

So, it forms 2 triangles  ABD and ACD,

Using the trigonometric ratios,

tan θ = Altitude / Base

DC = AD × tan (90° - x)

= 1600 × tan( 90° - 25.2° )

= 1600 × cot(25.2°)

= 1600 × 2.13

DB = AD × tan (90° - y)

= 1600 × tan( 90° - 71.7° )

= 1600 × cot(71.7°)

= 1600 × 0.33

∴ d = DC - DB

     = 1600 * [ tan( 90° - 25.2° ) - tan( 90° - 71.7° ) ]

    = 2,871 meters

To learn more about angle of depression from the given link

brainly.com/question/9723082

#SPJ4

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