hendrisyafa
17. sin 105 berada pd kuadran II yang nilainya sama dgn sin (180-105) = sin 75
sehingga : = (sin 15 - sin 75) (sin 15 + sin 75) = ( cos 75 - sin 75) ( cos 75 + sin 75) = cos²75 - sin²75 ---> cos²A - sin²A = cos 2A = cos (2.75) = cos 150 = - cos (180-150) = - cos 30 = - 1/2 √3
18. Mo = TbMo + (fa /fa+fb). c 56,25 = 55,5 + p-5 / (p-5+p-3) x 3 0,75 = (p-5) / (2p-8) x 3 3/4 = (p-5) / (2p-8) x 3 1/4 = (p-5) / (2p-8) 2p-8 = 4 (p-5) = 4p-20 2p-4p = -20+8 -2p = -12 p = 6
sin 105 berada pd kuadran II yang nilainya sama dgn sin (180-105) = sin 75
sehingga :
= (sin 15 - sin 75) (sin 15 + sin 75)
= ( cos 75 - sin 75) ( cos 75 + sin 75)
= cos²75 - sin²75 ---> cos²A - sin²A = cos 2A
= cos (2.75)
= cos 150
= - cos (180-150)
= - cos 30
= - 1/2 √3
18. Mo = TbMo + (fa /fa+fb). c
56,25 = 55,5 + p-5 / (p-5+p-3) x 3
0,75 = (p-5) / (2p-8) x 3
3/4 = (p-5) / (2p-8) x 3
1/4 = (p-5) / (2p-8)
2p-8 = 4 (p-5)
= 4p-20
2p-4p = -20+8
-2p = -12
p = 6
19. rata2 = 40/8 = 5
s =√ (7-5)²+ 2 (4-5)² + (1-5)² + 2 (5-5)² + (6-5)² +(8-5)² / (8-1)
=√ 4+2+16+0+1+9 / 7
= √ 32/7
≈ 2