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ruslelena [56]
3 years ago
14

Math help uwu ,,, will reward tyvm. (─‿‿─)

Mathematics
1 answer:
iren [92.7K]3 years ago
5 0

Answer:

Step-by-step explanation:

Here's how I would do it (not necessarily the only way or best way).

Draw an imaginary line from A to C.  This splits the kite into two triangles: ABC and ADC.  Since AB ≅ AD, CB ≅ CD, and AC ≅ AC, the two triangles must be congruent by SSS.  Therefore, ∠B ≅ ∠D.

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djyliett [7]

this is the answer to this question - Two weather tracking stations are on the equator 146 miles apart. A weather balloon is located on a bearing of N 35°E from the western station and on a bearing of N 23°E from the eastern station. How far is the balloon from the western station?

Answer:

Reasons:

The given parameters are;

Distance between the two stations = 146 miles

Location of the weather balloon from the Western station = N35°E

Location of the weather balloon from the Eastern station = N23°E

The location of the station = On the equator

Required:

The distance of the balloon from the Western station

Solution:

- The angle formed between the horizontal, and the line from the Western station

to the balloon = 90° - 35° = 55°

- The angle formed between the horizontal, and the line from the Eastern station

to the balloon = 90° + 23° = 113°

The angle at the vertex of the triangle formed by the balloon and the two stations is 180° - (55 + 113)° = 12°

By sine rule,

Distance from balloon to western station = 146/sin(12 dg) = Distance from balloon to western station/sin(113 dg)

Therefore;

Distance from balloon to western station = 146/sin(12 dg) x sin(113 dg) ~ 646.4

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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4 pairs of shorts,

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Answer: Mike descended 3000 feet

Step-by-step explanation: Hope this helps.

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