Points A, B, and C are collinear. Point B is between A and C. Find x if AB = 10x + 1, BC = 5x + 1, and AC = 17.
1 answer:
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Cos(x) = sin(x + 20o) Use the double cos angle formula
Cos(x) = sin(x)*cos(20) + cos(x)*sin(20) Divide through by cos(x). Tanx = sin(x)/cos(x)
1 = tan(x)*cos(20) + sin(20) Subtract sin(20) from both sides.
1 - sin(20) = tan(x)*cos(20) Calculate the value of 1 - sin(20)
0.65798 = tan(x) * 0.939692 Divide by cos(20)
tan(x) = 0.65795/0.939692
tan(x) = 0.70021 Take the inverse of 0.7) = tan(x)
x = tan-1(0.70021)
x = 35 degrees as expected.
Problem Two
The easiest way to do this is to pick random values and try it. The two acute angles are complementary. So 25 and 65 are random enough.
Let A = 25
Let C = 65
A
Tan(25) = sin(25)/sin(65) ; tan(25) = 0.4663.
sin(25)/sin(65) = 0.4663 Answer
B
Sin(90 - C) = Sin(A)
Tan(90 - A) = Tan(C)
So the question becomes Cos(A) = Tan(C) / sin(A)
Cos(25) = Tan(65)/Sin(25)
0.906 = ? 5.07 This statement isn't true.
C
Sin(65) = Cos(25)/Tan(65)
.906 = 0.4226 C is not the correct answer.
D
D isn't true. The tan does not relate that away. You can find it for yourself.
Cos(A) = Tan(C) You should get 0.906 = 2.14
E
Sin(C) = Cos(A) / Tan(A) I'll leave you to show this is wrong.
Problem 3
The diagram below is for this problem. Cos(x) = 50/100 = 0.5
x = cos-1(0.5)
x = 60 degrees.
Part 2
Sin(60) = opposite / hypotenuse
opposite = sin(60) * hypotenuse
opposite = 86.61 Be sure and round this to whatever the question says to round it to.
Answer:
0 ≤ Ф ≤ 4π.
Step-by-step explanation:
since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))
the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.
We take
Note that
Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.
The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.