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Trig Final Exam
供稿人: Burton
(原作者: Crouch)
Simplify cosx +sinxcotx

cosxsinx - cosxcotx

0

2cosx

1

Which of the following expressions is

equal to  1-sec4θ ?

-2tan2θ -tan4θ

tan2θ -tan4θ

2tan2θ -tan4θ

 -2tan2θ +tan4θ

        1

sinx + 1

-2cotxcscx

2sec2x

-2tanxsecx

2csc2x

Simplify
+

     1

sinx -1

Simplify

cot2x -1

cot2x +1

cannot be simplified

1-cot2x

tan2x -1

1-sec2x

Simplify (tanx-secx)(tanx+secx)
 1

-1

0

sec2x -tan2x

Solve 7 - 6sinx = 3 + 2sinx  for 0≤θ≤180ο

120ο

180ο

135ο

150ο or 30ο

Which of the following are the solutions

of  cot2x +2 =2cscx  on the interval [0,2π) ?

3π/2

0, π

π/2, 3π/2

π/2

Solve   2 cot2x - 4= 2

π/6 + nπ/2; 5π/6 + nπ/2

π/6 + nπ;   5π/6 + nπ

π/3 + nπ; 2π/3 + nπ

π/6 + 2nπ; 7π/6 + 2nπ

Solve....

0 ; 3π/2

0; π/2

π/2 

 ∏∕2, 3π/2

cotxsinx + cotx =0

Find the exact value of

tan155 + tans(-95)

1-tan155tan(-95)

1/√3

-√3

√3/3

√3

√6 - √2

         4

√6 + √2

          4

Find the exact value of cos15ο.

√6/4

√2

Find the magnitude of the horizontal

and vertical components of a velocity

of 23 mph at an angle of 16ο with the

ground.

horizontal: 191mph, vertical: 6mph

horizontal: 6mph, vertical: 22mph

horizontal: 22mph, vertical: 10mph

horizontal: 22mph, vertical: 6mph

A commercial passenger jet is flying with

an airspeed of 135 knots on a bearing

of N45οE.  If a 32-knot wind is blowing

at a bearing of S79οE,

determine the velocity and direction

of the jet relative to the ground.

120.1 knots, S55οE

129.2 knots, S35οE

155.2 knots, N55οE

155.2 knots, N35οE

Find the component form of AB with

intial point A(-8,2) and terminal

point B(2,9).

<6,12>

<10,7>

<11,12>

<11,6>

Find the magnitude of AB with initial

point A(-4,6) and terminal point

B(-1,1).

3.831

6.831

5.831

7.831

Given vectors

u=<-2,3> and  v=<-2,3>

<7,9>

<13,9>

<-8,12>

<-5,9>

Find 9u -5v

Find a unit vector u with the same direction

as x=<-40, 9>

<-9/41, 40/41>

 <-40/49, 9/49>

<-40/41, 9/41>

<-15/17, 8/17>

Find the direction angle of 13i + 15j.

56.65ο

229.09ο

139.09ο

49.09ο

An airplane is traveling due east with a

velocity of 563mph.  The wind blows at

76mph at an angle of S43οE.

What is the resultant speed and direction

of the plane?

617.3mph, S84.8οE

617.3mph, S5.2οE

620mph, S9.5οE

620.3mph, S85.2οE

Find the angle θ between

 u=<-4,-1> and v=<4,-4>

31ο

121ο

 149ο

59ο

Find one set of polar coordinates for

(-5, 5√3)

(10, 2π/3)

(5, 2π/3)

(2π/3, 10)

(-10, 2π/3)

Find the rectangular coordinates of

            (7, 30ο)

(-6.06, -3.5)

(6.06, -3.5)

(6.06, 3.5)

(-6.06,3.5)

(4, π/3)

(5, π/4)

(5, π/3)

(4, π/4)

Name the polar coordinates

of the point graphed.

A

How many triangles are there that satisfy

the conditions a=13, b=6 and A =61ο?

0

2

not possible

1

Given a triangle with a=14, A=41ο, and B=34ο
What is the length of c?
 21.6

20.6

19.6

22.6

5.3

28.1

14.1

10.6

What is the length of a?

Given a triangle with

b=7, c=9 and A=36ο

B
40ο
9
Solve ∆ABC
A
60ο
(round to the nearest degree or

nearest tenth in length)

C

<A =

a =
b =
ο
B
a=19, b=20, C=63ο

Determine whether ∆ABC should be solved

using Law of Sines of Law of Cosines.

A
C

Law of Cosines

Law of Sines

B
Solve......

A

19
(round to nearest degree or
nearest tenth in length)
63ο
20
C
<A =

<B =

c =
ο
ο
B
A=90ο, b=9, a=18

Determine whether ∆ABC should be solved

using Law of Sines of Law of Cosines.

A
C

Law of Cosines

Law of Sines

B
A=90ο, b=9, a=18

Solve ∆ABC.....

A
C
c=15.6, C=60ο, B=30ο
c=15.6, B=60ο, C=30ο
c=15.6, B=61ο , C=29ο 
c=15.6, C=61ο , B=29ο

Find the area of the triangle with

a=15ft, b=22ft and C=30ο.

142.9

82.2

165

285.8

  sq ft

sq ft

sq ft

 sq ft

Find the area of the triangle with

a=10ft, b=4ft and c=12ft.

15.7

18.7

20.7

19.7

 sq ft

sq ft

sq ft
sq ft
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