Question

No 1 sama 2 caranya gimana yaa??

Answer 1

1. *)Pusat (X)= (a,b)

= (-1/2A,-1/2B)

= (-1/2(-4),-1/2(8))

= (2,-4)

*)P(12,-4) adalah titik potong dengan garis singgung lingkaran berupa A dan B, maka jarak P ke pusat adalah:

|XP|² = (x-a)²+(y-b)²

|XP|² = (12-2)²+(-4-(-4))²

|XP|² = (10)²+(0)²

|XP|² = 100

|XP| = 10

*)L = AX . AP

kemungkinan:

48 = 6. 8

48 = 1. 48

48 = 2. 24

48 = 3. 16

48 = 4. 12

*) AX² + AP² = XP² -> karena tegak lurus

maka, dapat diketahui bahwa

6² + 8² = 10²

*) AX = r = 6

*) (x-a)²+(y-b)² -r² = 0

(x-2)²+(y-(-4))²-6² = 0

(x²-4x+4)+(y²+8y+16)-36=0

x²+y²-4x+8y-16=0

maka, c = 16

2. *)cos(x+25) = sa/mi = a

maka:

sa=a

mi=1

de=√(1²-a²) = √(1-a²)

*) sin(x+25) = de/mi = √(1-a²)/1 = √(1-a²)

misal x+25 = y

cos(2x+110)

= cos (2x+50+60)

= cos(2y+60)

= cos2y. cos60 - sin2y. sin60

= (2cos²y-1). cos60 - (2.siny.cosy). sin60

= (2a²-1). (1/2) - (2(√(1-a²))a).(1/2√3)

= (1/2)(2a²-1)-(a(√(1-a²))√3)

= (1/2)(2a²-1)-a√(3(1-a²))