Sagot :
Answer:
Step-by-step explanation:
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
Special Right Triangles:
We know one angle 30° and the opposite side of 5. Find BC & AB
Tan(30) = 5 / Adj ( where Adj = BC )
Adj = 5 / Tan(30)
BC = 8.660254038
BC = 8.7 ( rounded to nearest 10th )
AB = sqrt[ [tex]5^{2}[/tex] + [tex]8.660254038^{2}[/tex] ]
AB = sqrt[ 100 ]
AB = 10
next triangle
We know one angle , 45° and the adjacent side , 6 Find AC & BA
Tan(45) = AC / 6
6* Tan(45) = AC
6 = AC
BA = sqrt[ [tex]6^{2}[/tex] + [tex]6^{2}[/tex] ]
BA = sqrt [72]
BA = 8.485281374
BA = 8.5 ( rounded to nearest 10th )
We know that is triangle is an isosceles triangle, b/c of the right angle at the top, so the sides and bottom angles are the same, therefore
AB = 2*BD
AB = 8
CD = 3 ( b/c this is the famous 3,4,5 triangle)
AC = 5 ( b/c of symmetry)
Directed Segment:
We know the point C on the line AB is 1:2 along it and that point A is at (-4,6) and that point B is at (8,9) First find the distance between the points.
distance = sqrt [ (x2-x1)^2 + (y2-y1)^2 ]
where A = point 1 , P1 = (-4,6) in the form (x1,y1) and B = point2 , P2 in the form (x2,y2)
distance = sqrt [ (8-(-4) )^2 + ( 9-6)^2 ]
distance = sqrt [ 8 + 4 )^2 + (3^2 ]
distance = sqrt [ 12^2 + 9 ]
dist = sqrt [ 153 ]
dist = 12.36931688
ratio of 1:2 means 1/3 the distance along the line is where point C will be, from either end. that means there will be two different correct points for C to be at. Ask you teacher if they knew that. :/ the first correct point will be at 1/3* distance in x direction
1/3 * 12 = 4
in the y direction
1/3 *3 = 1
from point A in the x direction then it's -4+4 = 0 in the x direction
from point A in the y direction then it's 6+1 = 7
C1=(0,7)
the second point that would work is 2/3 along the line (b/c that would be 1/3 from point B or 1:2 ratio along the line )
from point B in the x direction then it's 8-4 = 4
from point B in the y direction then it's 9-1 = 8
C2 = (4,8)
either one of the points C1 or C2 will be correct for how this question was asked. ( in my opinion it was asked poorly , your teacher gets a D :0 )
Trig: ( refer to above helpful memory tool SOH CAH TOA for trig functions)
sin(A) = opp / Hyp
Sin(A) = 4 / 5
A = arcSin(4/5)
A= 53.13010235°
Cos(A) = Adj / Hyp
Cos(A)= 3/5
A = arcCos(3/5)
A = 53.13010235°
Tan(A) = Opp / Adj
Tan(A) = 4/3
A = 53.13010235°
Coordinate Geometry:
Find the midpoint of the line segment AB where A = (-2,-4) and B = (5,6)
midpoint = [(x1+x2)/2, (y1+y2)/2 ]
midpoint = ( -2+5 ) / 2 , ( 6 + (-4) )/ 2
midpoint = 3/2 , 2/2
midpoint = ([tex]\frac{3}{2}[/tex],1)
Find the length AB where A= (-2,3) , B= ( 6, -2)
dist = sqrt [ (x2-x1)^2 + (y2-y1)^2]
dist = sqrt [ 6-(-2))^2 + (-2-3)^2 ]
dist = sqrt [ 6+2)^2 + (-5)^2]
dist = sqrt [ 8^2 + 25 ]
dist = sqrt [64 + 25 ]
dist = sqrt [89]
dist = 9.43398
Pythagorean Theorem:
[tex]C^{2}[/tex] = [tex]A^{2}[/tex] + [tex]B^{2}[/tex]
(3, 4, 5) (5, 12, 13) (8, 15, 17)
Circle Equations:
circle at the origin and going thur a point P1 at (3,4)
x+h)^2 +y+k)^2 = [tex]r^{2}[/tex]
3^2 + 4^2 = [tex]r^{2}[/tex]
9+16 = [tex]r^{2}[/tex]
25 = [tex]r^{2}[/tex]
[tex]3^{2}[/tex] + [tex]4^{2}[/tex] = 25
circle at (x-4)^2 + (y+5)^2 = 81
centered at (h,k) = ( 4,-5)
radius of [tex]\sqrt{81}[/tex] = 9
circle of [tex]x^{2}[/tex] + [tex]y^{2}[/tex]+ 2x - 4y + 1 = 0
we have to complete the square for the x part and then the y part
( [tex]x^{2}[/tex]+2x +1 ) + ( [tex]y^{2}[/tex] -4y+[tex](-2)^{2}[/tex] ) = -1 +1 +4
(x+1)^2 + (y-2)^2 = 4
this circle is centered at (-1,2)
that took me a few hours.. send payment to... :D
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