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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:

This problem can be solved by a simple system of equations

I am going to call x the number of candy bars that Sally bought and y the number of sodas that Sally bought.

The fact that she bought 38 itens means the x + y = 38.

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