When graphing a trig function, the equation includes a a b h and k. In the equation y = 2sin(pi/2(x+1)+3, label each part and describe how it changes the equation.
a = 2 = amplitude
b = pi/2 = period
h = +1 = horizontal shift
k = +3 = vertical shift
cos(x)=1/3
Answer in radians
X= 1.231, 5.052
(look at paper)
Solve for θ in the given triangle
Angle A: ?
Angle B: 20 deg
Angle C: 126 deg
Angle A = 34 deg
Graph y = 4csc(x)+1
(see paper)
2sec(x)=-5
answer in radians
x= 1.982, 4.301
(look at paper)
Solve for side a in the given triangle
Angle A: 58 deg
Angle B: 71 deg
Angle C: 56 deg
Side AC: 19
side a = 16.05
Graph y = 6cot(1/2(x+pi))+2
(see paper)
5sin(1/2(x-pi))+3=0
x= 1.855, 10.712
(look at paper)
Solve for θ in the given triangle
Side AC: 38
Side AB: 44
Angle A: 30 deg
Angle B: ?
Angle B = 59.73 deg
Find equation from graph
y = 2tan(x)-1
-1/2cot(pi/6(x-1))+4=0
x= 1.238, 7.238
(look at paper)
Fully solve for the two possible triangles using the given information
Angle A: 25 deg
Side AC: 10
Side CB: 5
AB:?
Angle C?
Angle B:?
Triangle 1:
AB=11.74
Angle C=97.3 deg
Angle B=57.70 deg
Triangle 2:
AB=6.39
Angle C= 32.70 deg
Angle B=122.30 deg
Find equation and graph:
A ferris wheel:
total height: 586 m
wheel diameter : 446 m
one full revolution: 32 min
Then find height at 12 min
y = -223cos(pi/16(x)+363
height at 12 min: 537.21
1/4csc(pi/3(x+2))-1=0
Find 4 solutions
x= -1.759, 4.241
0.759, 6.759
+/- 6
(look at paper)
Find:
Angle ADC and side AC
Angle BDC and side DB
Given:
Angle DAC: 33 deg
Angle ACD: 78 deg
Angle ABC: 141 deg
Side AB: 102
Side BC: 76
AC=168.014
Angle ADC=91.46 deg
DB=106.13
Angle BDC=44.46 deg