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a person in a hot air balloon sights two cities, one behind other. The angles of depression of the cities are 45 degrees and 30 degrees. If the hot air balloon is 750 m high, how far apart are the two cities?

Sagot :

Let distance between two cities = x
Let C₂ = distance from farther city from the sight at 30 degrees
Let C₁ = distance from nearer city from the sight at 45 degrees
Height from the ground to air balloon = 750 m

Distance from below the balloon to the farther city (C₂):

tan 30⁰ = [tex] \frac{750}{C _{2} } [/tex]

C₂ = [tex] \frac{750}{tan30} [/tex]

C₂ = [tex] \frac{750}{0.58} [/tex]

C₂ = 1,293 m

Distance from below the balloon to nearer city (C₁):

tan 45 = [tex] \frac{750}{C _{1} } [/tex]

C₁ = [tex] \frac{750}{tan45} [/tex]

C₁ = [tex] \frac{750}{1} [/tex]

C₁ = 750 m

Distance  (x)between two cities:

x  =  C₂ - C₁ 

x = 1,293 m - 750 m

x = 543 m

ANSWER:  The distance between two cities are 543 meters.

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