Thursday

April 17, 2014

April 17, 2014

Number of results: 449,989

**Trigonometry (Law of Sine)**

Fire towers A and B are located 10 miles apart. They use the direction of the other tower as 0°. Rangers at fire tower A spots a fire at 42°, and rangers at fire tower B spot the same fire at 64°. How far from tower A is the fire to the nearest tenth of a mile?
*Saturday, February 25, 2012 at 4:03am by Alianne*

**Trigonometry (Law of Sine)**

Triangulation can be used to find the location of an object by measuring the angles to the object from two points at the end of a baseline. Two lookouts 20 miles apart on the coast spot a ship at sea. Using the figure below find the distance, d, the ship is from shore to the ...
*Saturday, February 25, 2012 at 4:04am by Alianne*

**Trigonometry (Law of Sine)**

Triangulation can be used to find the location of an object by measuring the angles to the object from two points at the end of a baseline. Two lookouts 20 miles apart on the coast spot a ship at sea. Using the figure below find the distance, d, the ship is from shore to the ...
*Saturday, February 25, 2012 at 4:09am by Alianne*

**Trigonometry (Law of Sine)**

Fire towers A and B are located 10 miles apart. They use the direction of the other tower as 0°. Rangers at fire tower A spots a fire at 42°, and rangers at fire tower B spot the same fire at 64°. How far from tower A is the fire to the nearest tenth of a mile? With solution ...
*Saturday, February 25, 2012 at 4:09am by Alianne*

**Trigonometry (Law of Sine)**

Your wording of the question is confusing, but I think I see a triangel FAB, where AB = 10 angle A = 42° and angle B = 64° then angle F = 180-64-42 = 74° FA/sin64 = 10/sin74 FA = 10sin64/sin74 = appr 9.4 miles
*Saturday, February 25, 2012 at 4:03am by Reiny*

**Trigonometry (Law of Sine)**

I see no "figure below", we cannot show diagrams in this forum. You will have to describe the figure.
*Saturday, February 25, 2012 at 4:09am by Reiny*

**Trigonometry (Law of Sine)**

see http://www.jiskha.com/display.cgi?id=1330160622
*Saturday, February 25, 2012 at 4:09am by Reiny*

**maths**

calculate the size of the angle or side marked on each of these diagrams
*Sunday, February 24, 2008 at 2:26pm by trigonometry*

****tRiGoNoMeTrY****

A weight is oscillating on the end of a spring. The position of the weight relative to the point of equilibrium is given by y=1/12(cos8t-3sin8t), where y is the displacement (in meters) and t is the time (in seconds). Find the times when the weight is at the point of ...
*Monday, November 17, 2008 at 10:46pm by *tRiGoNoMeTrY**

**math**

if sec theta = 2,what are the other five functions in values of theta?
*Wednesday, January 25, 2012 at 6:58pm by trigonometry*

**Matthew**

Given θ = arcsin (tan45°) find the exact degree measure of θ without usiing calculator.
*Tuesday, December 4, 2012 at 4:42pm by TRIGONOMETRY*

**Tyler**

Find the solution {in the interval of [0,2pi)} of: tan2x - 2cosx = 0 ( I know the answer is pi/6, but I guessed and checked. Please show the steps I need to go through in order to get the answer )
*Saturday, November 29, 2008 at 6:04pm by Trigonometry*

**Trigonometry**

What is your answer? Did you use the sine law?
*Monday, March 18, 2013 at 7:32pm by MathMate*

**Trigonometry**

I am curious how you find the probability of divisibility of palindromes by 4 using the sine law. Can you kindly show your work?
*Monday, March 18, 2013 at 7:32pm by MathMate*

**Trigonometry**

What are these things called: -cos x = cos -x What are the rules to these trig rules? Thanks
*Sunday, January 22, 2012 at 11:29pm by Trigonometry*

**Trigonometry**

Thanks for the prompt reply. How do these work out? The Odd/Even Identities? sin -x = -sin x cos -x = cox x tan -x = -tanx Why does this work out? Does it apply to certain quadrants or something? Thanks.
*Sunday, January 22, 2012 at 11:29pm by Trigonometry*

**trigonometry**

you are welcome little trick... when finding the angles in a triangle given the 3 sides, you will of course have to use the cosine law to find one angle, then you can use the sine law to find a second angle. Always find the largest angle first using the cosine law. That way if...
*Monday, March 19, 2012 at 7:29am by Reiny*

**trigonometry **

Use logarithms and the law of tangents to solve the triangle ABC, given that a=21.46 ft, b=46.28 ft, and C=32°28'30" I have to use logarithms and law of tangents . and then provide the check. which says law of sine or mollweids equation. I can't us the cosine law. If you ...
*Saturday, February 26, 2011 at 6:21pm by Anon*

**trigonometry ?**

A cottage under construction is to be 15.6m wide. The two sides of the roof are to be supported by equal rafters that meet at a 52 degree angle. Determine the length of the rafters to the nearest cm using a) the cosine law b) the sine law I am not understanding this question ...
*Tuesday, March 16, 2010 at 11:51am by deborah*

**Trigonometry **

By the cosine law 6.17^2 = 3.02^2 + 5.25^2 - 2(3.02)(5.25)cosA cosA = -.043279 angle A = 92.48 degrees I would then use the Sine Law to find the second angle, and then the "sum of the angles in a triangle" theorem to find the third one.
*Wednesday, December 9, 2009 at 2:50pm by Reiny*

**Trigonometry**

I think I did use the sine law but I am just not sure how to get a+b?
*Monday, March 18, 2013 at 7:32pm by Joe*

**Trigonometry**

You are obviously looking at or have made a diagram. let the top of the falls be point C In triangle ABC B = 52.9° A = 110.7° , the supplement of 69.3 then angle C = 16.4° by the Sine law AC/sin52.9 = 1000/sin16.4 AC = 2824.8913... now in the right-angled triangle sin69.3 = ...
*Saturday, June 16, 2012 at 10:39pm by Reiny*

**Trigonometry**

Make a sketch. I have a triangle ABC, where BC is the ground, AC is the tower. Angle B = 60°, angle C = 84.5°, making angle A = 35.5° by Sine Law: BC/sin 35.5° = 179/sin 60° I get BC = 120.026 So the shadow of the tower is 120 ft long
*Tuesday, August 30, 2011 at 10:55pm by Reiny*

**trigonometry **

I have posted a corrected version of solution using the law of tangents. I believe by now you are capable of doing the calculations (multiplications and divisions) using logarithms. I found the third side (c) using the cosine rule, but it gives the same answer as the sine rule...
*Saturday, February 26, 2011 at 6:21pm by MathMate*

**trigonometry**

If sine of an angle is ¼ and cosine of an angle is 15/4.. find cosecant.
*Saturday, March 7, 2009 at 8:45pm by Amber*

**trigonometry**

make a diagram label the tower AB , where A is the top of the tower. label the man's first position C, his second position D so that DC = 300 Look at triangle ADC, angle D = 35° angle ACD = 110° , so then angle DAC = 35° so the triangle is isosceles (lucky) and AC = 300 Then ...
*Tuesday, July 12, 2011 at 10:11am by Reiny*

**trigonometry**

The main part of the question is to make a decent diagram. We can find AC using Pythagoras and I found it to be 264.00 km Also using right-angled trig, AC makes an angle of 12.246° which makes angle PAC = 52.587° In triangle PAC , we have AP = 120 , AC = 264 adn angle PAC = 52...
*Saturday, January 5, 2013 at 11:51pm by Reiny*

**trigonometry **

If you know the sine of a number, how do you find its cosine?
*Sunday, December 16, 2012 at 5:06pm by rachel*

**Trigonometry**

There is no such thing. No angle can have a sine greater than 1. arcsin pi/2 would have to have a sine of pi/2. The sin of pi/2 is zero.
*Sunday, November 8, 2009 at 11:12pm by drwls*

**Trigonometry**

Cos D= -3/4. if the sine of the angle is positive, find the sine of the angle and determine the quadrant. Can anyone tell me the steps. I need to learn this.
*Thursday, February 18, 2010 at 7:02pm by Laly*

**trigonometry**

in triangle abc, m<c=90, bc=20, and ba=40. a.find sine a b.find measure of <a
*Tuesday, May 18, 2010 at 5:55pm by james*

**Trigonometry**

I made a side view diagram, with A and B the points on the lake, obviously A farther away from the tower. I labeled my tower DC , C as the base of the tower. In Triangle ABD, angle A = 7°, angle ABD = 167° , angle ADB = 6° , and AB = 1 km by the sine law: BD/sin7 = 1/sin6 BD...
*Wednesday, January 2, 2013 at 4:20pm by Reiny*

**trigonometry**

I made a sketch and got a triangle ABC where AB = 20 BC = 50 and angle B = 70° by the cosine law: AC^2 = 20^2+50^2-2(20)(50)cos70 = 2215.959... AB =47.07 m Using the sine law, find angle A
*Monday, April 23, 2012 at 10:14pm by Reiny*

**Trigonometry**

I made a sketch labeled end of first trip A, end 2nd trip B to get triangle OAB OA=4 , AB = 18 and angle BAO = 64° by simple geometry OA^2 = 18^2+4^2 - 2(4)(8)cos64 , by cosine law .. OA = 16.6395 m by sine law: Let angle AOB= Ø sinØ/18 = sin64/16.6395 ... Ø = 76.5 or 180-76.5...
*Tuesday, April 30, 2013 at 10:55pm by Reiny*

**trigonometry**

Construct the trapezoid ABCD ,where AB || CD AB = 12 and CD = 22 angle C=65 and angle D = 40 Draw AE || BC where E is on CD So now ABCE is a parallelogram, and CE = 12 which makes ED = 10 Now look at triangle AED, by corresponding angles angle AED = 65°, angle D = 40 leaving ...
*Friday, February 10, 2012 at 6:18am by Reiny*

**Trigonometry**

If the sine of an angle is 3/5 and the angle is not in quadrant 1, what are the remaining five trigonometric values for that angle?
*Sunday, January 24, 2010 at 7:19pm by Dan*

**Trigonometry**

If the sine of an angle is 3/5 and the angle is not in quadrant 1, what are the remaining five trigonometric values for that angle?
*Sunday, January 24, 2010 at 7:19pm by Dan*

**Trigonometry**

well, you know it's gotta be either sine or cosine of the angle. 0° means the whole weight, so it looks like 180cos10°
*Friday, November 15, 2013 at 12:41am by Steve*

**trigonometry**

make a sketch, label the side across from the 33º angle as x, and the side across the 62º as y. Using the sine law x/sin33 = 25/sin85 do y the same way. draw a perpendicular from the boat to shore, giving you a right-angled triangle, call it h sin 62 = h/x h = xsin62, and ...
*Monday, March 8, 2010 at 6:08pm by Reiny*

**Trigonometry**

Evaluate the sine, cosine and tangent of -7pi/3. I got 1/2, sq. root of 3/2 and sq. root of three. My book says the sine and the tangent are negative, but I don't know why? Can you explain the rule??
*Sunday, December 20, 2009 at 8:22pm by Sam*

**trigonometry**

in triangle abc, angle c is a right angle, AC=8,Bc=15 and AB=17 a.find Sine a b.Find cosine A c.Find Sine B d. Find COsine B E. the measure of angel A f. the measure of angela b Can u tell me the answers and how to do it. Because im in 8th grade
*Tuesday, May 18, 2010 at 5:58pm by james*

**trigonometry**

where is the sine negative? Isn't it in quadrants III and IV ? The sine of which angle is 1/2 ? Isn't it 30º or pi/6 radians so arcsin (-1/2) is 210º or 330º which would be 7pi/6 or 11pi/6 radians
*Tuesday, May 19, 2009 at 8:06pm by Reiny*

**trigonometry **

If you are not already familiar with the law of tangents, here's an article that can help you: http://en.wikipedia.org/wiki/Law_of_tangents We will be using the same notations in the following solution. In the given case, sides a,b are known, and the included angle C. So the ...
*Thursday, February 24, 2011 at 11:53pm by MathMate*

**Trigonometry**

Sine = Opposite/Hypotenuse --- SOH Cosine = Adjacent/Hypotenus -- CAH Tangent = Opposite/Adjacent -- TOA It helps you to memorize the definitions of the trig ratios in terms of the sides of a right-angled triangle
*Monday, September 27, 2010 at 8:32pm by Reiny*

**Trigonometry**

Law of sines: a/SinA = b/SinB solve for B from that. Then solve for C knowing A and B, and the sum of angles is 180 Then, find c with the law of sines, or law of cosines.
*Sunday, January 24, 2010 at 7:19pm by bobpursley*

**trigonometry**

how do you find sin theta and cosine theta of angles? ex. Find the sine theta and cosine theta of 150 degrees as well as the degrees in radians?
*Monday, February 4, 2008 at 7:02pm by mariya*

**Trigonometry**

I had to find the sine, cosine and tangent of -150 degrees. I got - sq. root of 3/2, -1/2, and - sq. root of 3/3. But my book has -1/2 as the sine, and -sq. root of 3/3 as the cosine. Why is this? I thought that it was thirty degrees away from the axis, so I should use pi/3.
*Sunday, December 20, 2009 at 8:25pm by Sam*

**Trigonometry**

express in terms of sine and cosine. a) tan theta/ cot theta i know the answer is sin^2 theta / cos ^2 theta but how do you do it?
*Thursday, March 22, 2012 at 5:45pm by Jonathan*

**Trigonometry**

When using the Law of Cosines, how do I know if two triangles exist?
*Tuesday, January 19, 2010 at 6:14pm by Laurianne*

**Trigonometry**

law of cosines b^2 = a^2 + c^2- 2ac cos B
*Monday, May 30, 2011 at 4:45pm by Damon*

**Trigonometry**

Solve the triangle using the law of cosine: a= 5 b= 6 y= 35
*Monday, July 8, 2013 at 11:12am by Hannah*

**Trigonometry law of cosine**

A plane is flying with an airspeed of 200 miles per hour and heading 150°. The wind currents are running at 30 miles per hour at 175° clockwise from due north. Use vectors to find the true course and ground speed of the plane. (Round your answers to the nearest ten for the ...
*Monday, April 7, 2014 at 3:41pm by NEED HELP ASAP*

**Trigonometry law of cosine**

draw the diagram. Add the wind vector to the speed vector (head to tail). The resultant from the origin to the head of wind is now the resultant. Label it R. You have two sides, and the angle between S and W. Figure it out. I see it as in my head as the 360-30-175 =155 check ...
*Monday, April 7, 2014 at 3:41pm by bobpursley*

**Trigonometry**

My diagram has the lighthouse as PQ with P at the top and PQ = 350 My ships are at A and B, angle at A = 4° and the angle at B = 6.5° In the right angled triangle BQP sin 6.5 = 350/BP BP = 350/sin 6.5 = ..... now look at triangle ABP we just found BP and angle ABP = 173.5° ...
*Monday, February 4, 2013 at 10:13pm by Reiny*

**Trigonometry**

You have no equations to solve, you probably meant "simplify the expressions" 1. sin^2 x (1/sin^2 x - 1) = 1 - sin^2 x = cos^2 x 2. (cosx/sinx) (1/cosx) = 1/sinx = cscx 3. recall the property of complementary trig ratios, that is... sin(π/2 - x) = cosx and cos(π/2 - ...
*Sunday, February 10, 2013 at 1:54am by Reiny*

** trigonometry**

okay so in my review packet under the "calculator" section it says law of sines and law of cosines. i dont know what that means. i wikipedia-ed it but i dont understand what they are saying. Please help!
*Thursday, January 22, 2009 at 8:50pm by Spencer*

**Trigonometry**

by law of cosines, c^2 = a^2 + b^2 - 2ab cosC Now, having a,b,c and C, use the law of since to get A,B. sinA/a = sinB/b = sinC/c
*Tuesday, March 18, 2014 at 9:40am by Steve*

**trigonometry**

Use a trig function that states opposite/hypotenuse. Of course, this would be the sine function. sin(45degrees) = h/7meters Let h = height of building sin45 = h/7 sin45 times 7 = h 4.949747468 = h Round 4.949747468 to the nearest tenths becomes 4.95 meters. The height of the ...
*Thursday, February 14, 2008 at 10:04pm by Guido*

**Trigonometry**

A right triangle with sides of 7 and 24 has a hypotenuse of 25. The cosine of arctan 7/24 is 24/25 and the sine is 7/25. Now use the formula cos2A = cos^2A - sin^A to arrive at cos(2arctan(7/24)) = (24/25)^2 - (7/25)^2 = 527/625
*Wednesday, November 12, 2008 at 6:44pm by drwls*

**Trigonometry**

You'll have to use the Ambiguous Case for the Law of Sines. I would give you a few links to some good websites. However, I'm not allowed. Try googling "ambiguous case law sines" without the quotes for some relevant information.
*Monday, November 26, 2007 at 10:40am by Michael*

**trigonometry**

on 1,2, use implicit diff, and then algebra to find dy/dx 3. draw the right triangle, you have two sides, the other you can calculate, and look at it, what is the sine? 4,5 use the definition of continuity. Perhaps you need to review that.
*Tuesday, June 4, 2013 at 11:29am by bobpursley*

**Trigonometry**

A triangle has two angles, one 90 degrees, the other (between adjacent and hypotamuse) 26 degrees, and a hypotamuse of 10ft. Find the opposite side's length. Hint: use sine. I think I'm supposed to do 10(sine26), but the answer isn't right. What am I missing? Thanks!
*Friday, September 6, 2013 at 8:59pm by DeeDee*

**trigonometry**

or sin (255°) = sin (180° + 75°) = - sin 75° [The terminal side of the angle lies in quadrant III, so the sine function is negative.] = - sin (45° + 30°) = - (sin 45°cos 30° + cos 45°sin 30°) [Use sin (A + B) = sin A cos B + cos A sin B.] = - [( Square root of (2) /2)( Square ...
*Wednesday, February 23, 2011 at 6:33am by anon*

**trigonometry**

How to verify this identity: 1+ tan x (over) = secant x sine x + cos x
*Wednesday, February 24, 2010 at 10:02pm by Debi*

**Trigonometry**

Using the Law of Cosines, how can I show that the measure of each angle of an equilateral triangle is 60 degrees?
*Monday, February 10, 2014 at 8:36pm by McKenna Louise*

**Trigonometry**

I'm trying to find the cosecant of 11 pi/6. The answer is -2, but I'm not sure where this came from. I thought You would flip the sine value which was sq. root of 3/2, making it 2/sq. root of three, which would then become sq. root of -6 over 3 or sq. root of -2. Why isn't ...
*Thursday, December 10, 2009 at 9:47pm by Delia*

**Trigonometry**

Find the 3 angles of the triangle, then use law of sines to find a, the side opposite angle A then, d = a sin34°
*Monday, February 4, 2013 at 10:24pm by Steve*

**Trigonometry**

by law of sines, 7.6/sin38 = 9.2/sinS knowing P+S+Q = 180 degrees, now you know Q all information about R is irrelevant
*Saturday, September 22, 2012 at 5:58pm by Steve*

**trigonometry**

For the sine and cosine functions, you need a calculator, a slide rule (if there is one in the attic) or a table of trigonometric functions. There are also web sites you can use. Sin 150 degrees = 1/2, exactly cos 150 degrees = -(sqrt 3)/2 You can prove that with geometry, but...
*Monday, February 4, 2008 at 7:02pm by drwls*

**trigonometry**

or, if you're not into vectors, try using the law of cosines and law of sines: the resultant r is r^2 = 2^2 + 12^2 - 2*2*12 cos120° r^2 = 172 r = 13.115 the angle θ between the big force and the resultant is sinθ/2 = sin120°/13.115 sinθ = 0.1321 θ = 7.6°
*Tuesday, November 27, 2012 at 12:58pm by Steve*

**Trigonometry**

In the equation: N=Asin(Bt) + C A represents half of the difference between the maximum and minimum, sometimes called amplitude. B is a multiplitive factor to ensure that the cycle of sine (2π) fits into the physical cycle, in this problem, 365 days. Since the amplitude of ...
*Monday, September 28, 2009 at 10:51pm by MathMate*

**trigonometry**

the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 degree and 40 degree respectively find the two legs. Answer using law of sines pls
*Friday, February 10, 2012 at 6:18am by joe*

**trigonometry**

In the old days when there were no calculators, surveying calculations like this would require logarithms to do the multiplications. There was a book about an inch and a half thick called "7-digit logarithm tables". We would look up sine(61° 2' 13" ) from the table, look up ...
*Sunday, February 20, 2011 at 11:30pm by MathMate*

**trigonometry**

using the law of sines, solve the rest of the triangle: side C=14 side B=4 angle c=145 degrees i got angle b= 9.4 angle a=25.6 side A=10.4 but i didn't know if it was correct
*Sunday, January 23, 2011 at 4:51pm by hw*

**Trigonometry**

a^2=b^2+c^2-2bcCosA CosA= 1/2 * (b^2+c^2-a^2)/bc A= 139 deg, you are right on that. Lets find angle B, by the law of sines sinB/b=SinA/a or sinB= .3269 or B=19 deg Lets find angle C by the law of sines SinC/c=SinA/a or SinC=.3718 C=21.8 deg check: do the sum of angles = 180 ...
*Tuesday, January 19, 2010 at 6:25pm by bobpursley*

**trigonometry**

If you draw this, you have a triangle with ASA. With those two angles, you can get the third angleC. Use the law of sines: a/SinA=8.68/sinC solve for side a. I agree with your answer.
*Friday, January 16, 2009 at 7:24pm by bobpursley*

**trigonometry**

let c = distance from A to top angle opposite c = 180 -36.67 = 143.33 angle oposite 526 = 180 - 143.33-26.5 = 10.17 law of sines sin 10.17/526 = sin 143.33/a so a = 1779 now sin 26.5 = h/1779 h = 794
*Wednesday, August 11, 2010 at 10:03am by Damon*

**trigonometry**

C = 180 - A - B = 180 - 45° 48' 36" - 61° 2' 13" = 73° 9' 11" Law of sines: a/sinA = c/sinC a = c.(sinA/sinC) = 426.(sin(45° 48' 36"))/sin(73° 9' 11")) a = 426*0.749186 a = 319.153
*Sunday, February 20, 2011 at 11:30pm by anon*

**trigonometry**

angle at ship B (sin 25)/20 = (sin A)/4.1 sin A = .0866 A = 4.97 deg angle at ship A A = 180 - 25 - 4.97 = 150.03 degrees then law of cosines d^2 = 20^2 + 4.1^2 - 2(20)(4.1)cosA
*Monday, January 6, 2014 at 7:43am by Damon*

**Trigonometry**

IM STUCK ON THESE :( 1. What is the equation for shifting the standard sine curve +2 units horizontally? A. y = sin (x + 2) B. y = sin x + 2 C. y = sin x − 2 D. y = sin (x − 2) 3. What is tan-Â¹ √3/3 ? A. π/4 B. -π/3 C. π/6 D. -π/4 4. cotâ€“1...
*Sunday, December 9, 2012 at 10:39am by Rodrick*

**Trigonometry**

IM STUCK ON THESE :( 1. What is the equation for shifting the standard sine curve +2 units horizontally? A. y = sin (x + 2) B. y = sin x + 2 C. y = sin x − 2 D. y = sin (x − 2) 3. What is tan-Â¹ √3/3 ? A. π/4 B. -π/3 C. π/6 D. -π/4 4. cotâ€“1...
*Monday, December 10, 2012 at 2:24pm by Erick*

**Trigonometry**

first you would want to convert feet to miles or vice versa. Then, youll want to build the triangle because you start at 40,000 feet and then that goes at a six degree angle along the x axis for 20 miles. So you will want to put this on a cartesian plane because from there you...
*Sunday, March 11, 2012 at 11:43pm by john*

**Trigonometry**

Hello, everyone: I am working on finding the exact values of angles that are less common and are therefor not found easily on the Unit Circle (at least, they are not labeled). For example, the problem I am asking about is: 10) Find the exact values of the Sine, Cosine and ...
*Monday, March 17, 2008 at 6:26pm by Timothy*

**trigonometry**

But how do I know where cosine (or tangent and sine) are positive? I don't understand. Thanks for the help, btw.
*Tuesday, May 19, 2009 at 8:06pm by Matt*

**trigonometry**

simple sine and cosine ratio problem sin 21 = depth/300 depth = 300sin21° = 10.75 m cos 21 = horizontal distance/ 300 hd = 300cos21 = 280.07 m
*Monday, December 10, 2012 at 5:52am by Reiny*

**trigonometry**

If you are trying to write sqrt(15), write it as ï¿½Ã£15, not 15ï¿½Ã£. A sine can not exceed 1, buy the way
*Monday, February 11, 2013 at 1:27am by drwls*

**trigonometry**

now use the cosine law to find AC
*Thursday, May 3, 2012 at 10:27pm by Reiny*

**Trigonometry**

for y = tan kØ , the period of the tangent curve is π/k (notice that this differs from the period definition for sine and cosine) so the period of tan (2x-π) is π/2 radians or 90° We know that tan (π/2) is undefined (a vertical asymptote) so 2x - π...
*Sunday, February 3, 2013 at 2:59am by Reiny*

**Trigonometry **

A ladder that is 15 feet long is placed so that it reaches from level ground to the top of a vertical wall that is 13 feet high A. Use the law of sines to find the angle that the ladder makes with the ground to the nearest hundredth B. Is more than one position of the ladder ...
*Tuesday, March 27, 2012 at 5:42pm by Melissa*

**Trigonometry**

one way is to use the law of cosines. a^2 = b^2+c^2 - 2bc cosA Without loss of generality, we may take a to be the smallest side, of length 2. Then, 2^2 = 4^2 + 5^2 - 2*4*5 cosA A = 22.33 similarly, B = 49.45 C = 108.22
*Tuesday, November 6, 2012 at 10:37pm by Steve*

**Trigonometry**

draw the picture as a at 3pm and at 5 pm. Label the distance the ship from the ship to the lighthouse as d. Note d^2=original distance^2+(15t)^2 where t=0 at 3pm, t=2 at 5pm, etc. Now, at t=2, using law of sines... original distance/sin(90-52)=d/sin90 so you know then d=...
*Saturday, June 23, 2012 at 10:31am by bobpursley*

**Trigonometry**

for the sin(A±B) and cos(a±B) for the Sine is goes sinAcosB ± cosAsinB and for Cosine is goes cosAcosB ... sinAsinB for the Sine, the signs stay the same, that is, sin(A+B) = sinAcosB + cosAsinB and sin(A-B) = sinAcosB - cosAsinB In the cosine formula, the signs are opposite, ...
*Monday, July 21, 2008 at 7:07am by Reiny*

**Trigonometry**

What do you mean by "solve"? You already have provided the three side lengths. The three angles A, B, and C can be computed by using the law of cosines. c^2 = a^2 + b^2 - 2 a *b cos C cos C = (a^2 + b^2 - c^2)/(2 a b) and two other rearrangements of the same formula for A and B
*Wednesday, January 12, 2011 at 7:23pm by drwls*

**trigonometry **

Use logarithms and the law of tangents to solve the triangle ABC, given that a=21.46 ft, b=46.28 ft, and C=32°28'30" Also Give checks. I know the use of logarithms is unnecessary but i have to show my solution by using logarithms and thats the only reason I'm asking for help. ...
*Thursday, February 24, 2011 at 11:53pm by Anon*

**Trigonometry**

Find all the solutions of the following triangle using the Law of Sines. Angle A: 83°20' Angle C: 54.6° c: 18.1 How would I find I find angle B? I know that you would subtract angle A and C from 180, but I do not understand how to maintain the minutes (') on my calculator.
*Thursday, December 8, 2011 at 5:01pm by Lindsey*

**Trigonometry**

these values right? tan theta=2/3 sec theta=Squareroot(13)/3 sine=2xSquareroot(13)/13 Cos=3xSquareroot(13)/13 Tan=2/3 Cosecant=Square root (13)/2 Secant=Squareroot(3)/3 Cotangent=3/2
*Sunday, January 10, 2010 at 6:39pm by Nick*

**Trigonometry**

tan= sin/cos Let C mean cosine, S mean sine(x) multiply by C 2C^2-3S=3 2(1-S^2)-3S=3 2S^2+3S+1=0 (2S+1)(S+1)=0 so you have the solution.
*Tuesday, April 12, 2011 at 11:22am by bobpursley*

**trigonometry**

your period is pi i got that by dividing 2pi(which is the regular period for cos and sine) by 2 and then you move everything up 2 and to the right pi
*Saturday, February 2, 2013 at 1:39am by vii*

**trigonometry**

sine is function 1/2cos^-1 is its angle. so i can convert cos^-1 into sin^-1 but can i cancel this with sin because there is (1/2).
*Tuesday, June 4, 2013 at 11:29am by edward*

**Trigonometry**

Use Law of Cosines: cosB = (a^2+c^2-b^2)/2ac. 0.58779 = (256+c^2-196)/32c 0.58779 = (c^2+60)/32c Cross multiply: 18.81c = c^2+60 c^2-18.81c+60 = 0. c = 4.1,and 14.7. Use Quadratic Formula.
*Wednesday, August 1, 2012 at 4:34pm by Henry*

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