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Naya [18.7K]
3 years ago
12

Please help!!!!!!! How many inches would a point on the outer edge of the large gear travel in a 150º rotation? (The gear has a

radius of 4 inches) Please explain how you would solve this.
Mathematics
1 answer:
damaskus [11]3 years ago
4 0
In a full rotation, 360°, a point on the edge of a gear would travel a distance equal to its circumference.  The circumference of a circle is:

C=2πr, and since we are only traveling 150°, we need to set up an appropriate ratio for circumference to the distance the point travels:

d/C=150/360

d=5C/12 and since C=2πr

d=10πr/12

d=5πr/12, and since r=4in

d=5*4π/12 in

d=20π/12 in

d=5π/3 in

d≈5.24 in (to nearest hundredth of an inch)
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3 years ago
From a window 20 feet above the ground, the angle of elevation to the top of a building across
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Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

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3 years ago
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ANSWER: 5 minutes 20 seconds

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Hope this helps brainliest and rate/thanks helps!

8 0
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