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belka [17]
2 years ago
6

Special Right Triangles

Mathematics
1 answer:
Leokris [45]2 years ago
7 0

Answer:

first option

Step-by-step explanation:

Using the sine/ cosine ratios in the right triangle and the exact values

sin30° = \frac{1}{2} , cos30° = \frac{\sqrt{3} }{2}

sin30° = \frac{opposite}{hypotenuse} = \frac{x}{32} = \frac{1}{2} ( cross- multiply )

2x = 32 ( divide both sides by 2 )

x = 16

----------------------------------------------------------

cos30° = \frac{adjacent}{hypotenuse} = \frac{y}{32} = \frac{\sqrt{3} }{2} ( cross- multiply )

2y = 32\sqrt{3} ( divide both sides by 2 )

y = 16\sqrt{3}

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_____

The attachment shows the original segment in red, the reflected segment in purple, and the rotated segment in blue. The equivalent line of reflection is shown as a dashed green line.

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