Tutors Answer Your Questions about Trigonometrybasics (FREE)
Question 994584: cos theta =  sqrt55/8 and 90 degrees < theta < 180 degrees. Find sin 2theta
Answer by ikleyn(988) (Show Source):
Question 994566: Prove that:
2cos^3 θcos θ/sinθ cos^2 θsin^3 θ= cot θ
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
Can't. It is only true at odd multiples of pi/2. Without the lead coefficient of 2 in the numerator, then it works. Was that a typo?
John
My calculator said it, I believe it, that settles it
Question 994411: The width of a rectangular is 6 ft less than 4 times the length. Write a model for the width W in terms of the length L.
Answer by macston(4006) (Show Source):
Question 994358: prove that : 1  sin theta / 1 + sin theta = ( sec theta tan theta)2
Answer by htmentor(912) (Show Source):
You can put this solution on YOUR website! (For easier typing, we will let theta = x)
If we multiply the numerator and denominator by 1  sin(x), and simplify we get:
(1  2sin(x) + sin^2(x))/cos^2(x) = (12sin(x))/cos^2(x) + tan^2(x)
Simplifying the first term gives
sec^2(x)  2sec(x)tan(x) + tan^2(x) = (sec(x)  tan(x))^2
Question 993812: You are supposed to show the steps it takes to get the answer:
(sec²x  6 tan x + 7)/ (sec²x  5) = (tan x  4)/ (tan x + 2)
Found 2 solutions by KMST, Alan3354: Answer by KMST(3791) (Show Source): Answer by Alan3354(47455) (Show Source):
Question 993808: You are supposed to show the steps it takes to get the answer:
(1 3cosx  4 cos²x)/ (sin²x) = (1 4cosx)/ (1 cosx)
Answer by Alan3354(47455) (Show Source):
Question 994020: from 1990 to 1998 the manufacturer's shipments for audio cassettes A (in millions) and compact discs C (in millions) can be modeled by the equations:
A=31.8t+322
c=428.8t+110
where t is the number of years since 1990. During what year did the number of compact discs shipped surpass the number of audio cassettes shipped?
Answer by Boreal(1464) (Show Source):
You can put this solution on YOUR website! The point where the two graphs meet is the year, if one looks at t.
31.8t +322=428.8t + 110
460.6 t= 212
t=0.460. This would be in mid1990. I wonder if the numbers were copied correctly, but with those equations, that is the value.
Question 993981: Find the solutions of the equation that are in the interval [0, 2pi;). (Enter your answers as a commaseparated list. If there is no solution, enter NO SOLUTION.)
22sint=2sqr(3)cost
Answer by Alan3354(47455) (Show Source):
Question 993889: A farmer has a triangular shaped field, FLD. DF350m, FL= 240m and DL=400m.The farmer wants to put up a fence f, and has exactly enough fence such that angle FXL=71 degrees. Calculate the length of fence f.
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! A farmer has a triangular shaped field, FLD. DF350m, FL= 240m and DL=400m.The farmer wants to put up a fence f, and has exactly enough fence such that angle FXL=71 degrees. Calculate the length of fence f.

Where is X ?
Question 993895: The diagram shows the location of airports P, Q and R. Airport Q is 130 km from Airport P on a bearing of 111 degrees. Airport R is 75 km from Airport Q on a bearing of 225 degrees.
a. Calculate angle PQR
b. Calculate the distance from Airport R to Airport P, giving your answer to the nearest whole number.
c. Calculate angle RPQ
d. Calculate the bearing of Airport P from Airport R.
Answer by Alan3354(47455) (Show Source):
Question 993896: The diagram shows the location of airports P, Q and R. Airport Q is 130 km from Airport P on a bearing of 115 degrees. Airport R is 75 km from Airport Q on a bearing of 225 degrees.
a. Calculate angle PQR
b. Calculate the distance from Airport R to Airport P, giving your answer to the nearest whole number.
c. Calculate angle RPQ
d. Calculate the bearing of Airport P from Airport R.
Answer by Alan3354(47455) (Show Source):
Question 993849: Sin3ASIN2A equal to zero then what is the value of A
Found 2 solutions by ikleyn, Alan3354: Answer by ikleyn(988) (Show Source): Answer by Alan3354(47455) (Show Source):
Question 993872: A farmer has a triangular shaped field, FLD. DF350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.
Found 2 solutions by ikleyn, Alan3354: Answer by ikleyn(988) (Show Source):
You can put this solution on YOUR website! .
A farmer has a triangular shaped field, FLD. DF350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.

There are 3 angles. Angle F is opposite the longest side.
Use the Cosine Law:
400^2 = 350^2 + 240^2  2*350*240*cos(F).
I got m(LF) =~ 83° (approximately).
For the Cosine Law see, for example, the lesson Proof of the Law of Cosines revisited in this site.
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! A farmer has a triangular shaped field, FLD. DF350m, FL= 240m and DL=400m. Calculate the angle to the nearest degree.

There are 3 angles. Angle F is opposite the longest side.
Use the Cosine Law:
400^2 = 350^2 + 240^2  350*240*cos(F)
Question 993803: You are supposed to show the steps it takes to get the answer:
sec²∅ + csc²∅ = sec²∅csc²∅
(∅= theta)
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! sec^2 + csc^2 = sec^2*csc^2
Multiply thru by sin^2*cos^2

sin^2 + cos^2 = 1
I'm convinced.
==========================
PS If anyone can post an Identity proof that shows a difference when working on one side only I would like to see it.
Question 993791: Find sin2b if sin b = 7/25
Answer by ikleyn(988) (Show Source):
Question 993815: You are supposed to show the steps it takes to get the answer:
(sin³A + cos³A)/ (sin A + cos A)= 1  sin A cos A
Answer by ikleyn(988) (Show Source):
Question 993701: I'm in pre calculus and I need help with solving a problem. It is y=cot(xpi/4) I need to find the period, frequency, phase shift, vertical shift, amplitude, and the asymptotes.
Answer by Theo(5548) (Show Source):
You can put this solution on YOUR website! your equation is y=cot(xpi/4)
below is the graph of the equation of y = cot(x) and y = cot(x  pi/4).
the graph of y = cot(x) is in orange.
the graph of y = cot(x  pi/4) is in black.
you will see that the graph of y = cot(x  pi/4) is shifted to the right of the graph of y = cot(x) by pi/4 radians.
the normal interval of the cotangent function is 180 degrees, or pi.
this means the graph repeats every 180 degrees, or every pi radians.
interval shown for y = cot(x) is 0 to pi.
interval shown for y = cot(x  pi/4) is pi/4 to 5pi/4.
the general form of a sinusoidal type wave is:
y = a * sin(b * (xc)) + d or:
y = a * cos(b * (xc)) + d or:
y = a * tan(b * (xc)) + d or:
y = a * cot(b * (xc)) + d or:
a is the amplitude
b is the frequency
c is the horizontal shift
d is the vertical shift.
sin and cos are sinusoidal type of wave with a period of 360 degrees because their pattern repeats every 360 degrees.
tan and cot are a sinusoidal type of wave with a period of 180 degrees because their pattern repeats every 180 degrees.
you want to find the period, frequency, phase shift, vertical shift, amplitude, and the asymptotes for the equation of y = cot(x  pi/4).
the period is 180 degrees, or pi radians.
the frequency is 1 because it is now shown.
the phase shift is the same as the horizontal shift which is pi/4.
the vertical shift is 0 because it is not shown..
the amplitude is 1 because it is not shown.
the asymptotes are at pi/4 and 5pi/4.
Question 993709: If cot 17 equals 3.2709 find tan 73
Found 3 solutions by MathTherapy, ikleyn, Boreal: Answer by MathTherapy(4047) (Show Source): Answer by ikleyn(988) (Show Source): Answer by Boreal(1464) (Show Source):
Question 993593: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Alan3354(47455) (Show Source):
Question 993591: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Alan3354(47455) (Show Source):
Question 993598: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Alan3354(47455) (Show Source):
Question 993602: 1.the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Alan3354(47455) (Show Source):
Question 993604: 1the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Alan3354(47455) (Show Source):
Question 993592: the shadow of a pole when the angle of elevation of the sun is 32° is 5m longer than the shadow of the pole when the angle of elevation of the sun is 48°. what is the length of the longer shadow of the pole?
Answer by Theo(5548) (Show Source):
You can put this solution on YOUR website! tan(48) = y/x
tan(32) = y/(x+5)
solve for y in both equation to get:
y = x * tan(48)
y = (x+5) * tan(32)
since they're both equal to y, the expressions on the right side of each equation must be equal to each other, so:
x * tan(48) = (x + 5) * tan(32)
simplify the right side of the equation to get:
x * tan(48) = x * tan(32) + 5 * tan(32)
subtract x * tan(32) from both sides of the equation to get:
x * tan(48)  x * tan(32) = 5 * tan(32)
simplify the left side of the equation to get:
x * (tan(48)  tan(32)) = 5 * tan(32)
divide both sides of the equation by (tan(48)  tan(32)) to get:
x = (5 * tan(32)) / (tan(48)  tan(32))
solve for x to get:
x = 6.432096215
that's your solution.
x + 5 must therefore be equal to 11.432096215
now that you know the value of x and x + 5, you can use that information to solve for y.
it should be the same in both cases.
if y = x * tan(48), then y is equal to 6.432096215 * tan(48) which is equal to 7.143566553.
if y = (x + 5) * tan(32), y is equal to 11.43209621 * tan(32) which is equal to 7.143566553.
y is the same in both cases, as it should be.
your solution is that the length of the longer shadow of the pole is 6.432096215 + 5 = 11.432096215
Question 993633: If sin A=3/5 then find (tan A+sec A)^2
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! If sin A=3/5 then find (tan A+sec A)^2

Since sin = y/r, y = 3 and r = 5

Then, x = sqrt[5^23^2] = sqrt(16) = 4

So, tan(A) = y/x = 3/4 and sec(A) = r/x = 5/4

Ans: (tan(A)+sec(A))^2 = (3/4+5/4)^2 = 2^2 = 4

Cheers,
Stan H.

Question 993622: Please help me out with this problem! Solve for all solutions of t on the interval [0,2pi)
22sin(t)=2sqrt( 3 )cos(t)
Answer by Alan3354(47455) (Show Source):
Question 993625: Please help!! Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)8sin(t)+1=0
Found 2 solutions by stanbon, Alan3354: Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Approximate, to the nearest 10', the solutions of the equation in the interval [0°, 360°).
sin^2(t)8sin(t)+1=0

Use quadratic formula:
sin(t) = [8+sqrt(644)]/2

sin(t) = [8+sqrt(60)]/2

sin(t) = 8+sqrt(15)

Comment:: Both values for sin(t) are out of the range of the sin function.

Ans: No solution.
Cheers,
Stan H.

Answer by Alan3354(47455) (Show Source):
Question 993471: Find all solutions of the equation in the interval [0,2pi)
2cos^2xcosx=1
Answer by solver91311(20879) (Show Source):
Question 993436: We are supposed to show the process that gets you to the answer (1) in the problem
csc²∅ cos²∅csc²∅= 1
(∅= theta)
Answer by Cromlix(3061) (Show Source):
You can put this solution on YOUR website! Hi there,
csc²∅ cos²∅csc²∅= 1
[csc²∅ = 1/sin^2∅]
.......
csc²∅ cos²∅csc²∅= 1
1/sin^2∅ cos²∅ 1/sin^2∅ = 1
1  cos²∅/sin^2∅ = 1
[1  cos²∅ = sin^2∅]
sin^2∅/sin^2∅ = 1
Hope this helps :)
Question 993170: A circle has a radius of 5ft. Find the length s of the arc intercepted by a central angle of 140°
Do not round any intermediate computations, and round your answer to the nearest tenth.
Answer by ikleyn(988) (Show Source):
Question 992753: Given that cotx=5/6 and x is in quadrant 2, find sin2x, cos2x,tan2x
Answer by ikleyn(988) (Show Source):
You can put this solution on YOUR website! .
Given that cotx=5/6 and x is in quadrant 2, find sin2x, cos2x,tan2x

cot(x) = > = > = . >
= .. (1)
+ = 1, (2)
Now substitute (1) into (2). You will get
. = 1, or
= = .
Hence, sin(x) = , cos(x) =  (taking into account that x is in the second quadrant).
Now, when you know sin(x) and cos(x) values, you can calculate the remaining functions.
Use the trigonometry formulas
sin(2x) = 2sin(x)*cos(x), = . . . ,
cos(2x) =  = . . . ,
and then tan(2x) = = . . . .
Complete these calculations yourself.
Good luck!
Question 992819: Given that tan θ − cot θ = 2,
find the possible values of tan θ, giving your answers in an exact form.
Hence solve the equation
tan θ − cot θ = 2
giving all values of θ between 0 and 360
I'm very confused how the questions are different? I really don't understand what the first question is asking me to do.
Answer by ikleyn(988) (Show Source):
Question 992746: Given that cotx=3/5 and x is in quadrant 2, find sin2x, cos2x,tan2x
Answer by ikleyn(988) (Show Source):
Question 992754: Given that sinx=3/5 and it is in quadrant 1. Find sin2x+cos2x
Answer by Cromlix(3061) (Show Source):
You can put this solution on YOUR website! Hi there,
If sin x = 3/5
cos x = 4/5
tan x = 3/4
..........
sin 2x = 2 sin x cos x
sin 2x = 2 (3/5)(4/5)
sin 2x = 24/25
..........
cos 2x = 2 cos^2 x  1
cos 2x = 2(4/5)^2  1
cos 2x = 2(16/25)  1
cos 2x = 32/25  1
cos 2x = 32/25  25/25
cos 2x = 7/25
..........
Hope this helps :)
Question 992757: if tan theta = 1 / root 7, show that (cosec ^2 theta  sec ^2 theta )/cosec^2 theta +sec^2 theta )=3/4
Answer by Cromlix(3061) (Show Source):
You can put this solution on YOUR website! Hi there,
if tan theta = 1 / root 7
cosec theta = root 8/1
sec theta = root 8/root 7
...........
(cosec ^2 theta  sec ^2 theta )/cosec^2 theta +sec^2 theta )
(root 8/1)^2  (root 8/root 7)^2/ (root 8/1)^2 + (root 8/root 7)^2
(8/1  8/7) / (8/1 + 8/7)
((56 8)/7 / ((56 + 8)/7
[48/7 / 64/7]
[ 48 x 7]/ [7 x 64)
Cancel 7's top and bottom
=48/64
Divide top and bottom by 16
= 3/4
Hope this helps :)
Question 991736: Find all solutions of csc x = 2, where 0 ≤ x ≤ 2π. Can you explain step by step how you got the answer.
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Find all solutions of csc x = 2, where 0 ≤ x ≤ 2π.

If csc(x) = 2, sin(x) = 1/csc(x) = 1/2

If sin = 1/2, the reference angle is pi/6

Since sin is negative in QIII and QIV
x = pi+(pi/6) = (7/6)pi
or x = (2pi)(pi/6) = (11/6)pi

Cheers,
Stan H.

Question 992125: Y=1/2x x is great than or equal to 0 x>/=0
Graph and calculate the values for the six trigonometric functions of the angle theta given in standard position, if the terminal side of theta lies on the given line.
Sin=
Cos=
Tan=
Cose=
Sec=
CoTan=
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! Y=1/2x x is great than or equal to 0 x>/=0
If x = 1, y = 1/2
Then r = sqrt[1+(1/2)^2] = sqrt(3/4) = sqrt(3)/2

Graph and calculate the values for the six trigonometric functions of the angle theta given in standard position, if the terminal side of theta lies on the given line.
Sin= y/r = 1/sqrt(3)

Cos= x/r = 2/sqrt(3)

Tan= y/x = 1/2

Invert sin, cos, tan to get csc, sec, and cotan

Cheers,
Stan H.

Cose=
Sec=
CoTan=
Question 992448: solve for x E[0,2pi] :2sin^2xsinx=0
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! solve for x E[0,2pi] :
========================
2sin^2xsinx=0
Factor::
sin(x)[2sinx1] = 0

sin(x) = 0 or sin(x) = 1/2

x = 0 or x = pi/6 or x = (5/6)pi

cheers,
Stan H.

Question 992389: Given sin t = 1/5, sec t<0, find cos (t) and cot (t).
I do not know how to find out if any of the functions (cos, cot, or sin) are negative or positive.
This is what I have:
y=1,, r=5, and x=2√6
cos(t)=(2√6)/5, cot(t)=1/(2√6)
I don't know where to go from here
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
Since secant is negative (given) and we know that cosine is the reciprocal of secant, then cosine must be negative as well. You have the correct magnitude for cosine, but your sign is incorrect. When you solved for the value of cosine, you had to take the square root of cosine squared. When you take the square root of both sides of an equation you have to consider both possible signs for the result. That is why the problem specified a negative secant  so that you would know which sign to attach to cosine at the point where you took the square root.
Since sine is positive and cosine is negative, we know from the identity that tangent must be negative, and therefore cotangent, the reciprocal of tangent must be negative as well.
Your value for cotangent is incorrect. Sine over cosine is tangent, so , and therefore
John
My calculator said it, I believe it, that settles it

Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785
