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# PLANE TRIGONOMETRY PART 3

1)FIND THE SINE, CONSINE, TANGENT, AND COTANGENT OF: a)122 DEGREES b)315 DEGREES 12' c)275 DEGREES 13' 37" d)193 DEGREES 41' 51"
2)USING EXACT VALUES, FIND THE NUMERICAL VALUE OF: a)sin 30 DEGREES cos 240 DEGREES +sin 210 DEGREES sin 300 DEGREES b)tan 225 DEGREES + tan (-45 DEGREES)
3)EXPRESS EACH OF THE FOLLOWING AS THE SAME FUNCTION OF A POSITIVE ACUTE ANGLE: a)cos 112 DEGREES 33' b)tan 310 DEGREES c)cot 138 DEGREES 15' 10" d)sin (-140 DEGREES)4)EXPRESS EACH AS A FUNCTION OF 0:a)sin (270 DEGREES + 0) b)cos (pi + 0) c)tan (810 DEGREES + 0) d)sin (0-180 DEGREES)5)FIND THE VALUE OF: a)log sin 62 DEGREES 22' 33" b)log cot 28 DEGREES 13' 17" c)log cos 125 DEGREES 15' 23" d)log tan 78 DEGREES 45' 50" 6)FIND THE ACUTE ANGLE A, TO THE NEAREST SECOND, WHEN: a)log cos A=9.12575 b)log sin A=9.91655 c)log cot A=0.11975 d)log tan A=0.063237)SIMPLIFY: sin(90 DEGREES+x)sin (180 DEGREES+x)cos(90 DEGREES+x)cos(180 DEGREES-x) 8)SOLVE THE RIGHT TRIANGLE BY LOGARITHMS:A=28 DEGREES 30', b=18.3

1)FIND THE SINE, CONSINE, TANGENT, AND COTANGENT OF:

a) sin 122o = 0.848 ; cos 122o = -0.530 ; tan 122o = -1.600 ; cot 122o = -0.625

b) note: 315o 12' = 315.2o

sin 315o 12' -0.705 = ; cos 315o 12' = 0.710; tan 315o 12' = -0.993; cot 315o 12' = -1.007

c) 275o 13' 37" =(NNN) NNN-NNNNsup>o

sin 275o 13' 37" = -0.996; cos 275o 13' 37" 0.091= ; tan 275o 13' 37" = -10.931 ; cot 275o 13' 37" = -0.915 ;

d) 193o 41' 51" =(NNN) NNN-NNNNsup>o

sin 193o 41' 51" -0.237 = ; cos 193o 41' 51" = -0.972; tan 193o 41' 51" = 0.244; cot 193o 41' 51" = 4.103;

2)USING EXACT VALUES, FIND THE NUMERICAL VALUE OF:

a)sin 30o cos 240o +sin 210o sin 300o = (1/2)*(-1/2) + (-1/2)(-√3/2)

= -(1/4) + (√3/4) = (√3-1)/4

b)tan 225o + tan (-45o) = (1) + (-1) = 0

3)EXPRESS EACH OF THE FOLLOWING AS THE SAME FUNCTION OF A POSITIVE ACUTE ANGLE:

Note: sin(a) = sin(180o - a); cos(a) = -cos(180o - a); tan(a) = -tan(180o - a); sin(-a) = -sin(a); cos(-a) = cos(a); tan(-a) = -tan(a)

a)cos 112o 33' = -cos 67o 27'

b)tan 310o = tan (-50o) = -tan 50o

c)cot 138o 15' 10" = -cot 41o44'50"

d)sin (-140o ) = -sin 140o = -sin 40o

4)EXPRESS EACH AS A FUNCTION OF 0:

I am not entirely sure what this is asking. The problems as stated already are functions of 0 technically, and you could write each answer in various different ways incorporating 0....

a)sin (270o + 0) = sin270ocos0 + cos270osin0

b)cos (pi + 0) = cos(pi)cos0 - sin(pi)sin0

c)tan (810o + 0) = (tan 810 + tan 0)/(1 - (tan 810)(tan 0))

d)sin (0-180o) = sin0cos180o - cos0sin180o

5)FIND THE VALUE OF:

a)log sin 62 DEGREES 22' 33" = -0.053

b)log cot 28 DEGREES 13' 17" = 0.270

c)log cos 125 DEGREES 15' 23" = No answer, because cos 125 is a negative number and you can't take log of a negative number.

d)log tan 78 DEGREES 45' 50" = 0.702

6)FIND THE ACUTE ANGLE A, TO THE NEAREST SECOND, WHEN:

There are no solutions to (a) or (b) as I understand the problem - as you can see in the work below, finding A is out of the domain range because cos of any angle could never equal 109.12575 - a number much much bigger than 1. Am I misreading the problem? I believe this is log10(cosA) = 9.12575. Is this correct??? Please email me back and we will see if we can figure this out.

a)log cos A=9.12575

109.12575 = cosA

A = cos-1 109.12575

A =

b)log sin A=9.91655

109.91655 = sinA

A = sin-1 109.1655

A =

c)log cot A=0.11975

100.11975 = cotA

A = cot-1 100.11975

A = 0.18939 = 0o 1' 8"

d)log tan A=0.06323

100.06323 = tan A

A = tan-1 100.06323

A = 49.156259 = 49o 9' 22"

7)SIMPLIFY: sin(90 o + x)sin(180 o + x)cos(90 o + x)cos(180 o - x)

= [sin(90 o)cos(x) + cos(90 o)sin(x)]* [sin(180 o)cos(x) + cos(180 o)sin(x)]*[cos(90 o)cos(x) - sin(90 o)sin(x)]*[cos(180 o)cos(x) + sin(180o)sin(x)]

= [cos(x) + 0]*[0 - sin(x)]*[0 - sin(x)]*[-cos(x) + 0]

= -sin2(x)cos2(x)

8)SOLVE THE RIGHT TRIANGLE BY LOGARITHMS:

A=28 DEGREES 30', b=18.3

B = 61.5o or 61o 30'

C = 900

Side a = 9.936

Side c = 20.823

Customer: replied 9 years ago.