𝑃𝑟𝑜𝑓𝑒𝑠𝑠𝑜𝑟 𝐸𝑑𝑔𝑎𝑟 𝑅 .𝑆 .𝑆𝑜𝑢𝑧𝑎
180 °=π 𝑟𝑎𝑑→5 ∙18012 →÷12
÷12
=75
𝑐𝑚 rad122𝜋2𝛼
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
Top Related