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satela [25.4K]
3 years ago
14

FERRIS WHEEL The Ferris wheel at an amusement park has a radius of 10.2 meters. A rider, who is sitting level with the center of

the Ferris wheel, is 13.1 meters above the ground. At what height above the ground is the rider when the wheel rotates another 140°? Round to the nearest tenth if necessary.
Mathematics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

19.7 m

Step-by-step explanation:

Draw a circle to represent the ferris wheel with a 10.2 m radius. Draw a horizontal line through the circle center horizontal to the ground (at 13.1 m). Now draw a line from the center of the circle (ferris wheel) upward at 140 degrees until you reach the edge of the circle. Draw a line down from that point to the horizontal line at This is your right triangle to solve.

Hypotenuse is 10.2 m (radius of circle)

The angle of the triangle the supplementary angle of 140 degrees. 180-140=40 degrees

sin 40 degrees=rise /radius=rise/10.2

sin 40=0.6428=rise/10.2

rise=10.2(0.6428)

rise=6.6 m

13.1+6.6=19.7 m

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Any number that is divisible by 3 is divisible by nine
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3 years ago
Read 2 more answers
Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me​
o-na [289]

Answer:

\frac{3}{2}

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

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

cos(120 + x) = cos120cosx - sin120sinx

                   = - cos60cosx - sin60sinx

                   = - \frac{1}{2} cosx - \frac{\sqrt{3} }{2} sinx

squaring to obtain cos² (120 + x)

= \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

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

cos(120 - x) = cos120cosx + sin120sinx

                   = -cos60cosx + sin60sinx

                   = - \frac{1}{2}cosx + \frac{\sqrt{3} }{2}sinx

squaring to obtain cos²(120 - x)

= \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

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

Putting it all together

cos²x + \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x + \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

= cos²x + \frac{1}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}(cos²x + sin²x) = \frac{3}{2}

                 

5 0
3 years ago
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