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A hot air balloon is rising vertically. From a point on level ground 110 feet from the point directly under the passenger compartment, the angle of elevation to the balloon changes from 25 degrees to 32 degrees. How far, the nearest tenth, does the balloon rise during this period

Sagot :

Answer:

17.44 ft.

Step-by-step explanation:

Given,

The Distance of the hot air balloon: 110 ft

Change in elevation, [tex]\theta_1 = 25^0, \theta_2 = 32[/tex]

We know that,

[tex]\tan \theta = \dfrac{Perpendicular}{base}[/tex]

Now,

[tex]\tan \theta_1 = \dfrac{h_1}{110}[/tex]

[tex]\tan 25^0= \dfrac{h_1}{110}[/tex]

[tex]h_1 = 51.294\ ft[/tex]

[tex]\tan \theta_2= \dfrac{h_2}{110}[/tex]

[tex]\tan 32^0= \dfrac{h_2}{110}[/tex]

[tex]h_2 = 68.736\ ft[/tex]

Rise of balloon = 68.736 - 51.294

                         = 17.442 ft

                         = 17.44 ft

Hence, the rise of the balloon rise is 17.44 ft.

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