𝑃𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝐸𝑑𝑔𝑎𝑟 𝑅 .𝑆 .𝑆𝑜𝑢𝑧𝑎

180 °=π 𝑟𝑎𝑑→5 ∙18012 →÷12

÷12

=75

𝑐𝑚𝑔𝑟𝑎𝑢𝑠122𝜋2𝛼

𝛼=2∙2𝜋12 ¿

4𝜋  12 ¿

𝜋3

1

3

≅𝜋2 𝑟𝑎𝑑

¿𝜋2 𝑟𝑎𝑑 rad

¿𝜋2 𝑟𝑎𝑑

𝑜𝑏𝑠 :2𝜋𝑟𝑎𝑑 ≅ 6,28𝑟𝑎𝑑

𝑒𝑛𝑡 ã 𝑜 𝜋2 𝑟𝑎𝑑≅ 1 ,57𝑟𝑎𝑑

𝜋4 5𝜋

4

9𝜋4 13𝜋

4

(√22 ; √22 )(− √22; √22 )

(− √22;− √2

2 ) (√22 ;− √22 )

x

𝑇𝐴𝐵1x

𝑇𝐴𝐵2

𝑘=2

2

𝑃 𝑡

𝑡=1∙2𝜋  4 𝜋

𝑡=12

Logo 𝑘− 𝑡=2− 12=4−12 =

32

1

(0,2)

𝑓 (0 )=1−1=0𝑓 (0 )=1+1=2𝑓 (0 )=cos (0−1)=cos (−1)𝑓 (0 )=cos (0+1)=𝑐𝑜𝑠1

𝐶 (𝑥)

𝐿 (𝑥 )=𝑉 (𝑥 )−𝐶 (𝑥)

V ( x)

𝐿 (𝑥 )=𝑉 (𝑥 )−𝐶 (𝑥)¿3−2=1→𝐿 (𝑥 )=1000.000

1

24

1

(3,3)

𝝅𝟐

!!!

𝑡𝑔𝛼=𝐶𝑂𝐶𝐴 → 𝑡𝑔50 °= 𝑃𝑂𝑁𝑇𝐸

10 →1,19= 𝑃𝑂𝑁𝑇𝐸10

→𝑃𝑂𝑁𝑇𝐸=10 ∙1,19→𝑃𝑂𝑁𝑇𝐸=11,9𝑚

𝐶𝑎𝑡𝑒𝑡𝑜𝑜𝑝𝑜𝑠𝑡𝑜𝐶𝑎𝑡𝑒𝑡𝑜𝑎𝑑𝑗𝑎𝑠𝑐𝑒𝑛𝑡𝑒

200𝑚1,5𝑚

h 𝐴𝑙𝑡=h+1,5𝑚

𝑡𝑔𝛼=𝐶𝑂𝐶𝐴→ 𝑡𝑔30 °= h

200→0,577=h200→h=200 ∙0,577

→h≅ 115,4𝑚 𝐴𝑙𝑡=h+1,5𝑚→ 𝐴𝑙𝑡=115,4+1,5→ 𝐴𝑙𝑡≅ 116,9𝑚