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Vitek1552 [10]
3 years ago
10

A hot air balloon is released into the air. During its straight ascent, the angle of elevation was 15° and, 3 minutes later, the

angle of elevation increased 20°. How fast is the balloon traveling, in km/h, if the angle measurements were taken 300m away from the launch site?
Mathematics
1 answer:
Leni [432]3 years ago
7 0

Answer:

The speed of balloon is 0.16 m/s.

Step-by-step explanation:

BC = 300 m

AD = x

AB = y

time, t = 3 min = 3 x 60 = 180 s

By the triangle, ADC,

tan 15 = \frac{AD}{BC}\\\\0.27 \times 300 = x \\\\x = 80.4 m

By the triangle, BDC

tan 20 = \frac{BD}{BC}\\\\0.36 \times 300 = x + y \\\\x + y = 109.2 m

So, y = 109.2 - 80.4 = 28.8 m

Speed of balloon = 28.8/180 = 0.16 m/s

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Step-by-step explanation:

Solution:

- The sum of the distances from a point on the ellipse to its foci is constant. You have both foci and a point, so you can find the sum of the distances.

-Then you can find the vertices since they are points on the ellipse on the x-axis whose sum of distances to the foci are that value.

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                                     c^2 = a^2 + b^2

Where, a: x-intercept

            b: y-intercept

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                                     b^2 = 7*a

- Substitute latus rectum expression in the first one we get:

                                    c^2 = a^2 + 7a

                                    a^2 + 7a - 144 = 0

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                                    a = +/- ( 16 )

- The y-intercept we will use latus rectum expression again:

                                    b = +/- √(7*16)

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- The intercepts are:

                                    x - ( 16 , 0 ) , ( -16 , 0 )

                                    y - ( 0 ,  4√7 ) , ( 0 , -4√7)  

                                     

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