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3 years ago

Please help me with this trigonometry question:

Mathematics

1 answer:

zzz [600]3 years ago

8 0

Answer:

The distance from A to B is 736.2 to the nearest tenth foot

Step-by-step explanation:

In ΔCAB

∵ m∠CAD = 30° ⇒ exterior angle of Δ at vertex A

∴ m∠CAD = m∠ACB + m∠ABC

∵ m∠ABC = 20°

∴ m∠ACB = 30° - 20° = 10°

We will use the sin rule to find the distance AB

∵ $\frac{sin20}{1450}=\frac{sin10}{AB}$

∴ $AB=\frac{1450(sin10)}{sin20}=736.1842936$ ≅ 736.2 to the nearest tenth foot

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BartSMP [9]

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3 years ago

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lora16 [44]

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Now, we find the area of the figure:

We have that by definition, the area of a rectangle is given by:

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Where:

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Answer:

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3 years ago

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sammy [17]

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3 years ago