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sammy [17]
3 years ago
11

La figura MNPQ es un rombo. Si 2θ – Φ =60°. Calcula el valor de “x”. Dime los fundamentos y principios que utilizas.

Mathematics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

x = 30°

Step-by-step explanation:

Donde 2θ - Φ = 60 °, tenemos;

∠QMN = ∠QPM

Ф + ∠P / 2 + (180 - θ) = 180 (Suma de ángulos en un triángulo ΔQLP)

Por lo tanto, Ф + ∠P / 2 - θ = 0

Por lo tanto, ∠P / 2 = θ - Ф, también

270 ° + θ + x + ∠P / 2 = 360 (Suma de ángulos en un NOLP cuadrilátero) da

270 ° + θ + x + θ - Ф = 270 ° + 2 · θ - Ф + x = 360

Como 2 · θ - Ф = 60 °, tenemos;

270 ° + 60 + x = 360

x = 360 - 270 - 60 = 30 °

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Step-by-step explanation:

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<em>Additional comment</em>

This geometry is impossible, because the height from the long side cannot be more than the length of the short side. Here, the short side is 12.5 units, so it is not possible for the height to be 23 units.

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