Monday

April 21, 2014

April 21, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**trig**

Mass = 725Lbs * 0.454kg/Lb = 329.2 kg. m*g = 329.2 * 9.8 = 3226 N. = Wt. of trailer. Fp = 3226*sin36 = 1896 N. = Force parallel to the ramp. F = Fp = 1896 N. Required.
*Thursday, April 17, 2014 at 8:55pm*

**Trig**

thank you
*Thursday, April 17, 2014 at 12:38am*

**Trig**

use law of cosines. 70^2 = 65^2 + 90^2 - 2(65)(90)cosθ
*Thursday, April 17, 2014 at 12:29am*

**trig**

a 725 pound trailer is sitting on a ramp inclined at 36 degrees. how much force is required to keep the trailer from rolling down the ramp??
*Wednesday, April 16, 2014 at 10:25pm*

**Trig**

Force 1= 65 pounds , Force 2=90 pounds find the angle between the forces? magnitude of the resultant force= 70lbs
*Wednesday, April 16, 2014 at 10:11pm*

**Trig**

ARggghhhh!!!!
*Wednesday, April 16, 2014 at 5:37pm*

**Trig - heh**

now, just change all those 225's to 255's and redo the calculations.
*Wednesday, April 16, 2014 at 5:57am*

**Trig**

we need cos 225° and sin225° 225 = 180+45 , it is in quadrant III and has the same numerical trig values as 45°, but in III both the sine and cosine are negative sin 225 = - sin 45° = -1/√2 cos 225 = - cos 45° = -1/√2 so 7(cos255degrees + ...
*Tuesday, April 15, 2014 at 8:52pm*

**Trig**

how do you find the standard form of 7(cos255degrees + isin255degrees) ?
*Tuesday, April 15, 2014 at 8:08pm*

**Geometry- trig ratio**

25,636 ft?
*Tuesday, April 15, 2014 at 7:46pm*

**Geometry- trig ratio**

sin 55 = 21,000/hypotenuse we want that hypotenuse
*Tuesday, April 15, 2014 at 7:44pm*

**Geometry- trig ratio**

An observer (o) spots a plane flying at a 55 degrees angle to his horizontal line of sight. if the plane is flying at an altitude of 21,000 ft, what is the distance (x) from the plane (p) to the observer (o)?
*Tuesday, April 15, 2014 at 7:38pm*

**Trig**

A vertical pole stands on a 20° slope from 50 m down the slope from the pole the angle of elevation of the top of the pole is 35 degrees. How tall is the pole
*Tuesday, April 15, 2014 at 5:40pm*

**Trig**

A vertical tower is on a 12° slope. from a 50 m down the hill, the angle of elevation of the top of the tower is 30°. find the height of the tower
*Tuesday, April 15, 2014 at 4:03pm*

**HELP!**

Next year (sophomore year) I'm taking the following courses: Honors Algebra II/Trig Spanish III Health AP Biology Honors English AP World History Is AP Bio and world history hard? or no it's not just the exam. Please tell me because this is my first year taking AP.
*Monday, April 14, 2014 at 8:00pm*

**trig**

A car is on a driveway that is inclined 12 degrees to the horizontal. A force of 470lb is required to keep the car from rolling down the driveway. a) Find the weight of the car b) Find the force the car exerts against the driveway
*Monday, April 14, 2014 at 3:44pm*

**trig**

How did you get 2x=211 ?
*Monday, April 14, 2014 at 9:20am*

**trig/ vectors**

he needs to head upstream at an angle θ (measured from the line CD), such that tan θ = 4/12 so, θ = 18.4° The speed required is √(16+144) = 12.65 mph
*Sunday, April 13, 2014 at 11:46pm*

**trig/ vectors**

Points C and D are directly across from each other on opposite banks of a river. A boat travels across the river directly from point c to point d at a speed of 12 mph. If the current of the river has a speed of 4 mph, at what angle, and speed, must the captain head to travel ...
*Sunday, April 13, 2014 at 8:46pm*

**trig**

no
*Friday, April 11, 2014 at 11:43pm*

**trig**

cos(x) assumes all values between -1 and 1. Each of those values appears twice in a period of the curve. Since you want a unique value, x = 0 so cos(x) = 1 or x = 180, so cos(x) = -1 for any other value where cos(x) = k, cos(360-x) is also equal to k. Look at the graph and you...
*Friday, April 11, 2014 at 10:55am*

**trig**

just use your calculator to find the angle with the given property. Then solve for x. For example, tan 58.7° = 1.6443 Since the period of tan(x) is 180°, you have to find x where 90-2x = 58.7: 2x = 31.3, so x = 16 2x = 211, so x = 106 also, continuing on, x = 196,286 ...
*Friday, April 11, 2014 at 10:52am*

**trig**

Given angle x, where 0 <= x < 360 (degrees), cos(x) is equal to a unique value. Determine the value of to the nearest degree. Justify your answer.
*Friday, April 11, 2014 at 10:50am*

**trig**

Steve - Could you elaborate on your answer please? I don't understand where you got those two numbers from
*Friday, April 11, 2014 at 10:46am*

**trig**

Given angle x,where 0 <= x <= 360 (degrees) solve for to the nearest degree. a)cos(2x) = 0.6420 b)sin(x + 20) = 0.2045 c)tan(90 - 2x) = 1.6443
*Friday, April 11, 2014 at 10:42am*

**trig**

either 1 or -1 is attained only once on the given domain.
*Friday, April 11, 2014 at 10:41am*

**trig**

Given angle x, where 0 <= x < 360 (degrees), cos(x) is equal to a unique value. Determine the value of to the nearest degree. Justify your answer.
*Friday, April 11, 2014 at 10:39am*

**trig**

Given angle x, where 0 <= x < 360 (degrees), cos(x) is equal to a unique value. Determine the value of to the nearest degree. Justify your answer.
*Friday, April 11, 2014 at 10:38am*

**trig**

thanks
*Thursday, April 10, 2014 at 5:46pm*

**trig**

cosA = 3/5 sinB = 15/17 tanA = 4/3 tanB = 15/8 tan(A-B) = (tanA-tanB)/(1+tanAtanB) = (4/3 - 15/8)/(1+(4/3)(15/8)) = (-13/24)/(7/2) = -13/84
*Thursday, April 10, 2014 at 5:36pm*

**trig**

A and B are positive acute angles. if sin A=4/5 and cos B=8/17 find the value of tan (A-B) is the answer =43/100
*Thursday, April 10, 2014 at 5:26pm*

**trig**

540 @ 50° = <413.66,347.11> wind is <0,-27> add them up to get <413.66,320.11> which is 523 @ 37.7°
*Thursday, April 10, 2014 at 5:21pm*

**trig**

An airplane has air speed of 540 mph and a heading of 50*. The wind is blowing from the north at 27 mph. Find the plane's ground speed and course direction.
*Thursday, April 10, 2014 at 3:57pm*

**trig**

consider the identity Sin(A+B)= then, look at the identties for double angles when A is 2Theta
*Thursday, April 10, 2014 at 9:16am*

**trig**

Use standard identities to express sin(2θ+π/3) in terms of sinθ and cosθ
*Thursday, April 10, 2014 at 8:20am*

**trig**

arc tan Theta= 50000/30000 theta = arc tan ( 5/3 )
*Wednesday, April 9, 2014 at 7:14pm*

**trig**

a plane is 50000 ft in the air and 30000 ft from an airport find the angle of depression from the plane to the airport
*Wednesday, April 9, 2014 at 7:02pm*

**algebra 2/trig**

1/x + 1/8 = 1/3 x = 24/5
*Wednesday, April 9, 2014 at 11:41am*

**algebra 2/trig**

Tina can paint a room in 8 hours, but when she and her friend Emily work together, they can complete the job in 3 hours. How long would it take Emily to paint the room alone?
*Wednesday, April 9, 2014 at 10:34am*

**algebra**

Im using this for trig functions ....so I cannot use fraction or decimal I have to simply use π. Im getting : π-3/2 (2)π-3/2 ***I don't know how to add 2π +(-3)? thank you for helping
*Tuesday, April 8, 2014 at 7:50pm*

**trig**

cscx+1/cscx cosx = secx+tanx
*Tuesday, April 8, 2014 at 5:06pm*

**maths Pls help trig**

but oops: 25/√50 = 5/√2, not 1/√2
*Tuesday, April 8, 2014 at 1:31pm*

**maths Pls help trig**

the denominator is 50√50
*Tuesday, April 8, 2014 at 1:30pm*

**maths Pls help trig**

Thank u so much !! But I still have a question in#1 where did the 50 go?
*Tuesday, April 8, 2014 at 1:00pm*

**MathS triG**

see solution above of the same question Please don't switch names.
*Tuesday, April 8, 2014 at 11:34am*

**maths Pls help trig**

1. tanA = 1/3, and tanB = 1/7 I usually construct my triangles, (since the fraction gives you 2 of the sides of right-angled triangles, the third can always be found using Pythagoras) for tanA = 1/3, sinA = 1/√10 , cosA = 3/√10 for tanB = 1/7, sinB = 1/√50, ...
*Tuesday, April 8, 2014 at 11:33am*

**maths Pls help trig**

1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA-2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[-360;360] is the identity in 2.1 undefined? 3.If tanB=3/4 , 0<B<90 , prove that 4cos2B+3sin2B=4 4.Prove ...
*Tuesday, April 8, 2014 at 11:01am*

**MathS triG**

1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA-2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[-360;360] is the identity in 2.1 undefined? 3.If tanB=3/4 , 0<B<90 , prove that 4cos2B+3sin2B=4 4.Prove ...
*Tuesday, April 8, 2014 at 10:51am*

**trig**

You lean a ladder 6.7 m long against a wall. It makes an angle of 63 degrees with the level ground. How high up is the top of the ladder?
*Monday, April 7, 2014 at 7:41pm*

**Math - Int Trig**

I'll do one. Do the others in like wise. If you get stuck, come on back with what you did. (g∘h) = 3/(h^2+2) = 3/(x-2+2) = 3/x (g∘h)(-2) = 3/-2 = -3/2 The domain of h is x>=2 The domain of g is all reals, since x^2+2 is never zero. So, the domain of (g∘...
*Thursday, April 3, 2014 at 7:47pm*

**Math - Int Trig**

I can't edit but f(x) is suppose to be f(x)=3x+2 I put 2x by mistake!
*Thursday, April 3, 2014 at 7:46pm*

**Math - Int Trig**

Given the functions: f(x)=2x+2 g(x)=3/x^2+2 h(x)=sqrt(x-2) Find the following 1) (f∘g)(-5) 2) (h∘g)(3) 3) (g∘h)(-2) 4) (g∘g)(-2) 5) (f∘h)(12) Find the following and state the domain: 6) h∘g 7) f∘g 8) g∘f 9) g∘g 10)g∘h
*Thursday, April 3, 2014 at 7:43pm*

**trig**

Thanks a lot.
*Wednesday, April 2, 2014 at 9:35pm*

**trig**

after 1 hours, one ship has gone 25 km, the other 15 km make a sketch, let the distance between them be x straigh-forward case of the cosine law x^2 = 25^2 + 15^2 - 2(25)(15)cos 60° = .... you do the arithmetic. (I got appr 21.8 km)
*Wednesday, April 2, 2014 at 7:18pm*

**trig**

Help! Ships A and B leave port at the same time and sail on straight paths making an angle of 60 degrees with each other. HOw far apart are the ships at the end of 1 hour if the speed of ship A is 25 km/h and that of ship B is 15 km/h?
*Wednesday, April 2, 2014 at 6:18pm*

**not - trigonometry**

Why are you calling this trig ? assuming his first deposit is NOW end of 1st year: 1000(1.06) + 1000 = 2060 end of 2nd year: 2060(1.06) + 1000 = 3183.60 ... end of 5th year: .......... = 6637.09 fill in the rest using your calculator
*Tuesday, April 1, 2014 at 9:23pm*

**trig**

I assume you want the distance traveled along the hypotenuse, d, then sin30° = 5000/d d = 5000/sin30 d = ... if you want the horizontal distance traveled , x tan30 = 5000/x x = 5000/tan30 = ...
*Tuesday, April 1, 2014 at 6:10pm*

**trig**

a hot air balloon climbs continuously along a 30° angle to a height of 5,000 feet. To the nearest tenth of a foot, how far has the balloon traveled to reach 5,000 feet? Draw a sketch and then solve.
*Tuesday, April 1, 2014 at 5:51pm*

**trig**

I drew a circle, and a vertical radius of 12 m In 40 s, it rotates 360°, so in 5 sec it rotated 45° let the vertical position from the centre be y cos45 = y/12 y = 8.485 m so he is 12-8.85 or appr 3.5 m high after 5 seconds
*Monday, March 31, 2014 at 10:27am*

**trig**

The ferris wheel in an amusement park. It has a diameter of 24m, and it takes 40s to make one complete revolution. If Peter gets on a gondola which is vertically below the centre of the ferris wheel, find his rise in the height after 5s.
*Monday, March 31, 2014 at 8:54am*

**Trig**

see the graphs here: http://www.wolframalpha.com/input/?i=plot+y%3Dcos2x+and+y%3D-2sinx However, it might be better to graph y = -cos2x, then the final graph would be easier to visualize: http://www.wolframalpha.com/input/?i=plot+y%3D+-cos2x+and+y%3D-2sinx%2C+y+%...
*Sunday, March 30, 2014 at 6:26am*

**Trig**

y=0 is not the y-axis. And anyway, in the stated domain, x=pi is the axis of symmetry.
*Sunday, March 30, 2014 at 6:22am*

**Trig**

y = 0 The graph of y = cos x is symmetrical about the y axis.
*Sunday, March 30, 2014 at 2:02am*

**Trig**

If 0 ≤ x ≤ 2π, which equation is a line of symmetry for the graph of y = cos x? x = 0 x = 2π x = π y = 0
*Sunday, March 30, 2014 at 1:04am*

**Trig**

On the same set of axes, sketch and label the graphs of the equations y = cos 2x and y = –2 sin x in the interval 0 ≤ x ≤ 2π. How many values of x in the interval 0 ≤ x ≤ 2π satisfy the equation –2 sin x – cos 2x = 3? A.1 B.2 C....
*Sunday, March 30, 2014 at 1:03am*

**Trig**

sinC/c = sinA/a
*Saturday, March 29, 2014 at 10:52pm*

**Trig**

you have two solutions for tanA, one positive, one negative. So, there are two values for A, one in each QI, one in QII.
*Saturday, March 29, 2014 at 10:51pm*

**Trig**

period = 1/freqeuncy
*Saturday, March 29, 2014 at 10:49pm*

**Trig**

In ΔABC, a = 19, c = 10, and m∠A = 111. Which statement can be used to find the value of C?
*Saturday, March 29, 2014 at 9:52pm*

**Trig**

What is the total number of solutions for the equation 3 tan^2 A + tan A – 2 = 0 in the interval. 0 ≤ A ≤ π?
*Saturday, March 29, 2014 at 9:51pm*

**Trig**

If the period of a cosine function is 1/3 what is the frequency of the function?
*Saturday, March 29, 2014 at 9:50pm*

**trig**

I meant 336.7°
*Thursday, March 27, 2014 at 11:25am*

**trig**

Assuming the normal 0° at North, and ignoring the misuse of the word "bearing", I'd say that the resultant heading of the plane is 246.7°
*Thursday, March 27, 2014 at 11:24am*

**trig**

Not sure what notation you are using when you say "blowing in at 53° " is North your 0° and you are going clockwise, or is 0° your positive x - axis and you are going counterclockwise ?
*Thursday, March 27, 2014 at 10:34am*

**trig**

A plane is flying due south at a speed of 192 mph. a wind is blowing in at 53 degrees at 13 mph. what is the bearing of the plane?
*Thursday, March 27, 2014 at 10:09am*

**Trig Identities**

If hypotenuse is 1 opposite = sin T adjacent = cos T 1^2 = (sin T) ^2 + (cos T)^2 as we know and sin^2 T = 1 - cos^2 T sin T = +/- sqrt (1-cos^2 T) well, that is not much of an identity if the sign is arbitrary. BUT if we use the negative root then we know that cos^2 T is ...
*Tuesday, March 25, 2014 at 7:25pm*

**Trig Identities**

Explain why Sin of theta= -The square root of 1-cos^2 theta is not an identity, using either graphical or numerical reasoning.
*Tuesday, March 25, 2014 at 6:40pm*

**Pre-Calc**

don't forget your algebra I just because you're doing trig now. 2cos2θ = √3 cos2θ = √3/2 Now it's time to recall where cos(x) is positive: QI and QIV 2θ = π/6 or -π/6 Now, cos(x) has period 2π, so cos(2x) has period π...
*Tuesday, March 25, 2014 at 5:07am*

**trig**

tan^2(5x)cos^4(5x) = 1/8 - 1/8cos(20x) since tan = sin/cos, you have sin^2(5x)cos^2(5x) = 1/8 - 1/8cos(20x) Now use the double-angle formula on the left, and the half-angle formula on the right to get 1/4 sin^2(10x) = 1/4 sin^2(10x) QED
*Tuesday, March 25, 2014 at 12:14am*

**trig**

I'd help, but can't figure out (tx At any rate, use the identity tan^1+1 = sec^2 to reduce everything to a polynomial in cos(x). Solve for cos(x) and then for the values of x.
*Tuesday, March 25, 2014 at 12:10am*

**trig**

tan^2(5x)cos^4(5x) = 1/8 - 1/8cos(20x)
*Monday, March 24, 2014 at 8:32pm*

**trig**

tan^2(5x)cos^4(tx = 1/8 - 1/8cos(20x)
*Monday, March 24, 2014 at 8:32pm*

**trig**

sin T = 714/1000 so csc T =1/sin T= 1000/714 = 500/357 tan T = 714/700 = 102/100= 51/50 cos T = 7/10 You have them right but mislabeled.
*Monday, March 24, 2014 at 1:49pm*

**trig**

Find csc(theta), tan (theta), and cos (theta), where theta is the angle shown in the figure. Give exact values, not decimal approximations c=10 b=7 a=7.14 the right angle is locate between sides a and b and the theta angle is an acutle angle sides b and c. i have sin theta as ...
*Monday, March 24, 2014 at 1:29pm*

**Trig - sum and difference formulas**

the important identity here is cos(A+B) = cosAcosB - sinAsinB So, it looks like you have cos(6x+x) = cos 7x You also know that sin(A+B) = sinAcosB + cosAsinB so, sin(90+x) = sin90cosx + cos90sinx = 1*cosx = 0*sinx = cosx Do lots of these, since they will come in handy later on.
*Monday, March 24, 2014 at 5:13am*

**Trig - sum and difference formulas**

Write the expression as a single trigonometric functions: cos 6x cosx - sin6x sin x if you could do it step by step? also... How do I prove that sin(90deg + theta) = cos(theta)?? also step by step so i may learn?
*Monday, March 24, 2014 at 12:21am*

**Trig**

sinx = 2cosx tanx = 2 tanx > 0 in QI or QIII, so there are two values for x. Verify this at http://www.wolframalpha.com/input/?i=solve+sinx+%3D+2cosx
*Sunday, March 23, 2014 at 7:49pm*

**Trig**

The number of values in the interval –π ≤ x ≤ π that satisfy the equation sin x = 2 cos x is 1 2 3 4 0
*Sunday, March 23, 2014 at 6:46pm*

**trig**

sin(anything) lies between -1 and +1 so -18sin(71π t) lies between -18 and +18 then p lies between 83-18 and 83+18 or between 65 and 101 period of your function = 2π/(71π) = 2/71 min or appr 1.69 seconds 1.69 seconds = 1 beat 1 second = 1/1.69 beats 60 seconds...
*Sunday, March 23, 2014 at 8:36am*

**trig**

Suppose that Magnus' blood pressure can be modeled by the following function {{{p(t)=83-18sin(71pi*t)}}} Magnus' blood pressure increases each time his heart beats, and it decreases as his heart rests in between beats. In this equation, p(t)is the blood pressure in ...
*Sunday, March 23, 2014 at 7:05am*

**trig**

okay, thanks!
*Thursday, March 20, 2014 at 12:51am*

**trig**

tan(kx) has period pi/k, since x grows k times faster. So, I think you have a typo. It should be 3tan3x = 3 tan3x=1 now, you know that tan pi/4 = 1, so 3x = pi/4, and x = pi/12. Now, tan(3x) has period pi/3. 2tanx-2 = 0 tanx = 1 Since tan pi/4 = 1, tan (pi/4 + pi) = tan 5pi/4...
*Thursday, March 20, 2014 at 12:32am*

**trig**

I have an equation that is 3tan2x=3 and the solution is x=pi/12 + npi/3 and another equation that is 2tanx-2=0 , the answer sheet says the solution is x=5pi/4. why is the npi not added to the second solution but it is on the first?
*Thursday, March 20, 2014 at 12:20am*

**trig**

as you said, sin and cos have period of 2pi and tan has period pi. That's how you know.
*Thursday, March 20, 2014 at 12:07am*

**trig**

cos/sin use 2npi and tan uses npi, but how do you know when to add 2npi or npi to the solution of the equation?
*Wednesday, March 19, 2014 at 9:32pm*

**math trig**

a^2 + 100 = 225 a^2 =125 = 25 * 5 a = 5 sqrt 5
*Wednesday, March 19, 2014 at 7:07pm*

**math trig**

cos/(1+sin) + (1+sin)/cos cos^2/[ cos(1+sin) ] +(1+sin)^2 /[cos(1+sin)] (cos^2 + 1 + 2 sin + sin^2)/[cos(1+sin)] but cos^2 + sin^2 = 1 ( 2 + 2 sin) /[cos(1+sin)] 2 (1+sin)/[cos(1+sin)] 2/cos which is 2 sec
*Wednesday, March 19, 2014 at 7:06pm*

**math trig**

a triangle has a measured of hypoteneuos 15 and an adjacent of 10 solve for the triangle.
*Wednesday, March 19, 2014 at 7:03pm*

**math trig**

verify the identities COSX OVER 1PLUS SINX PLUS 1PLUS SINX OVER COSX IS EQUAL TO 2SEC.
*Wednesday, March 19, 2014 at 7:00pm*

**math trig**

a triangle abc has a mearsure of an adjacent 15 and 25 degree solve for the triangle
*Wednesday, March 19, 2014 at 6:54pm*

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