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April 19, 2014

April 19, 2014

**Recent Homework Questions About Trigonometry**

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**Trig**

Don't know. I supplied both solutions, one using the pythagorean theorem, and one using the law of cosines. pick one, and tell me the first place you get stuck. I'll try to walk you through it. I prefer the law of cosines, since it's for a trig class. If you don&#...
*Tuesday, October 16, 2012 at 6:35pm*

**Trig**

They aren't explained in a way that I can understand them. Can you explain more clearly?
*Tuesday, October 16, 2012 at 6:21pm*

**Trig**

see related questions below
*Tuesday, October 16, 2012 at 6:16pm*

**Trig**

A baseball diamond is a square with sides 22.4m The pitchers mound is 16.8m from home plate on the line joining home plate and second base. How far is the pitchers mound from first base?
*Tuesday, October 16, 2012 at 6:14pm*

**trig**

two trains leave Kansas City at the same time. Train A is traveling due north at 55 mph, train B is traveling west at the rate of 65 mph. Find the distance between the two trains two hours later and the bearing of train B from train A.
*Tuesday, October 16, 2012 at 5:51pm*

**Calculus 1**

i am having problems with the trig identities that and how to solve with sin squared. i have the formula. that hasn't helped me with the identities.
*Tuesday, October 16, 2012 at 5:34pm*

**Math Trig**

if the hypotenuse is at (12,5), the hypotenuse is length 13: sinθ = 5/13 cosθ = 12/13 tanθ = 5/12 secθ = 1/cosθ = 13/12 cscθ = 1/sinθ = 13/5 cotθ = 1/tanθ = 12/5
*Tuesday, October 16, 2012 at 11:12am*

**Math Trig**

plot the point (12,5) and join it to the origin. draw a vertical line from the point to the x-axis You now have a right-angled triangle with a base (x) of 12 and a height (y) of 5 Using Pythagoras, we can find the hypotenuse h^2 = 12^2 + 5^2 = 169 h = 13 let the base angle be...
*Tuesday, October 16, 2012 at 10:40am*

**Math Trig**

I know the answer but I don't understand the process. ! If you could explain the process without the answers?!
*Tuesday, October 16, 2012 at 10:10am*

**Math Trig**

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (12, 5) lies on its terminal side? Please explain how!
*Tuesday, October 16, 2012 at 9:56am*

**Math Trig**

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (12, 5) lies on its terminal side? Please explain how!
*Tuesday, October 16, 2012 at 9:56am*

**TRIG**

If you are using standard notation, then a bearing of 53° would be 53° clockwise from North make a sketch. Start the plane at the origin, then draw a line 53° form the y-axis (NOrth) and make it 2.5(550) or 1375 miles long. Complete the right-angled triangle, whose...
*Tuesday, October 16, 2012 at 9:38am*

**TRIG **

An airplane flying at 550 miles per hour has a bearing of 53°. After flying for 2.5 hours, how far north and how far east will the plane have traveled from its point of departure? I'm having a REALLY hard time with this one. I don't know what I'm doing wrong. ...
*Tuesday, October 16, 2012 at 12:17am*

**TRIG.**

Make a sketch showing the steeple on top of the church. label the height of the church c and the height of the steeple as s I see two right-angled triangles: in the first: tan 35° = c/50 c = 5tan35 = appr 35.01 m in the second: (c+s)/50 - tan 47.33333.. c+s= 54.248 so the ...
*Monday, October 15, 2012 at 11:19pm*

**Math**

We don't need Pythagorus, we need trig. let the height be h h/5 = tan 72.5° h = 5tan 72.5 = appr 15.858 m
*Monday, October 15, 2012 at 11:10pm*

**TRIG.**

From a point 50 feet in front of a church, the angles of elevation to the base of the steeple and the top of the steeple are 35° and 47° 20', respectively. Find the height of the steeple. I missed class this day and have no idea how to go about this problem. I'...
*Monday, October 15, 2012 at 11:01pm*

**Trig**

Lol! Apparently I have been working on this question too long! Thanks!
*Monday, October 15, 2012 at 4:29pm*

**Trig**

plugging in the trig values, you have .1051x = .2309(1.1-x) Have you forgotten your algebra I? .1051x = .2540 - .2309x .3360x = .2540 x = 0.756
*Monday, October 15, 2012 at 4:22pm*

**Trig**

Yes I understand the diagram points and can see how you came up with this now but am having trouble coming up with an answer for xtan6°=(1.1-x)tan13° I know this is basic but I am having trouble isolating the x! I am getting x=tan13°-tan6°/1.1 which is giving ...
*Monday, October 15, 2012 at 4:18pm*

**Trig -- oops**

oops. Sorry. I see that AB=1.1 So, adjust the equations and solve again.
*Monday, October 15, 2012 at 4:00pm*

**Trig**

x is the distance from A to the base of the tower. Since AB=1, the distance to B is 1-x. Did you not understand the diagram points?
*Monday, October 15, 2012 at 3:59pm*

**Trig**

Can you show me the steps you used to isolate x?
*Monday, October 15, 2012 at 3:54pm*

**Trig**

Let the base of the tower be C, and the top of the tower be T. So, the height h=CT, and we have AB=1, so AC=x and BC=1-x h/x = tan6° h/(1-x) = tan13° so, solve for x: xtan6° = (1-x)tan13° x = 0.687 h = 0.0722km = 72.2m
*Monday, October 15, 2012 at 3:43pm*

**Trig**

From an observation tower that overlooks a runway, the angles of depression of point A, on one side of the runway, and point B, on the oppisite side of the runway are 6 degrees and 13 degreess respectivel. The points and the tower are in the same vertical plane and the ...
*Monday, October 15, 2012 at 3:27pm*

**Calc**

y=uvw y'=vwu' + uvw'+uwv' It is just algebra, and probably some trig identities.
*Monday, October 15, 2012 at 11:22am*

**trig - incomplete**

If the plane is at 12000', then the cliff height h is found using h = 12000 + 4000 sec14° = 16122
*Monday, October 15, 2012 at 10:08am*

**trig**

A plane is flying 12,000 feet horizontally from a tall, vertical cliff. The angle of elevation from the plane to the top of the cliff is 45degrees, while the angle of depression from the plane to a point on the cliff at elevation 8000 feet is 14degrees. Find the height of the ...
*Monday, October 15, 2012 at 3:02am*

**trigonometry**

amazing. I didn't think anyone did spherical trig any more.
*Friday, October 12, 2012 at 10:15am*

**Trig**

if the inverse is y = 5/(x+4) , then the inverse of the inverse (the original) is x = 5/(y+4) xy + 4x = 5 y = (5-4x)/x so we have f(x) = (5-4x)/x and g(x) = 4(x-2) g(5) = 4(5-2) = 12 f(g(5)) = f(12) = (5-48)/12 = -43/12
*Friday, October 12, 2012 at 8:46am*

**Trig**

Let f and g be two invertible functions such that f^-1(x)=5/x+4 and g(x)=4(x-2). Find f(g(5)). Show your steps please so I can see how to do it. Thank you! :)
*Friday, October 12, 2012 at 1:44am*

**Calculus 2**

I don't understand why you were instructed to use trig substitution for this question, it is straightforward You should recognize certain pattern of derivatives and integrals Notice that the derivative of the denominator is 2x and we have x at the top, so this follows the ...
*Wednesday, October 10, 2012 at 9:35pm*

**trig**

Make a sideview sketch you should have 2 triangles, one right-angled containing the height and a scalene triangle with angles 24° , 153° (the supplement of 27°) and 3° the side opposite the 3° angle is 1000 by let the side opposite the 24° be x, (also ...
*Wednesday, October 10, 2012 at 9:16pm*

**trig**

A survey team is trying to estimate the height of a mountain above a level plain. From one point on the plain, they observe that the angle of elevation to the top of the mountain is 24. From a point 1000 feet closer to the mountain along the plain, they find that the angle of ...
*Wednesday, October 10, 2012 at 7:50pm*

**Precalc/Trig**

0<x<(π/2) ---> x must be in quadrant I 1/√3 < cotx < √3 consider the "ends" cotx = 1/√3 ---> tanx = √3 ---> x = π/3 or (60°) cotx = √3 ---> tanx = 1/√3 ---> x = π/6 or (30°) so &#...
*Wednesday, October 10, 2012 at 7:46pm*

**Precalc/Trig**

Find all values of x, 0<x<(pi/2) for which (1/( root 3))<cotx<(root 3)
*Wednesday, October 10, 2012 at 7:22pm*

**Math/Trig**

The period is 2pi/b. So when you plug it in, 2pi/1/2 would result in 2pi x 2/1. In the end it would be 4pi.
*Wednesday, October 10, 2012 at 6:08pm*

**trig**

1 i think
*Wednesday, October 10, 2012 at 5:14pm*

**trig**

csc(x) = 1/sin(x) so, what's 1/1?
*Wednesday, October 10, 2012 at 4:32pm*

**trig**

what is sin x = 1, what is the value of csx
*Wednesday, October 10, 2012 at 4:29pm*

**trig**

since pi/3 = 60 degrees, it should be easy to figure the above values.
*Wednesday, October 10, 2012 at 5:11am*

**trig**

what is 7pi/3 in degree .13pi/3 in degree
*Wednesday, October 10, 2012 at 1:44am*

**Trig.**

Never mind, I figured it out.
*Wednesday, October 10, 2012 at 1:29am*

**Pre-Calculus-Trig**

A. 3617.6 B. 4018.7 C. 3642.12
*Tuesday, October 9, 2012 at 9:30pm*

**Trig.**

I'm trouble solving for theta. Any help is appreciated. 0.73377324 = sin(theta)/cos(theta)
*Tuesday, October 9, 2012 at 6:22pm*

**trig**

Find the domain of each function. f(x)=Sqareroot of 5-x f(x)=x^3/x^2-16
*Tuesday, October 9, 2012 at 6:15pm*

**trig**

Find the following values for each function a. f(0) b. f(-1) c. f(-x) d. f(x+1) f(x)=2x^2+3x-4 I you plug in the value for each x, I just want to check my answers. Thanks.
*Tuesday, October 9, 2012 at 6:13pm*

**Trig help?**

yes no , http://www.wolframalpha.com/input/?i=cos%28x%5E2%29 yes , http://www.wolframalpha.com/input/?i=cos%28sin%28x%29%29
*Tuesday, October 9, 2012 at 12:20pm*

**Trig help?**

are these functions periodic? i - y=abs(sin(x)) = sin(x) ii - y=cos(X^2) iii - y = cos(sin(x)) I am completely lost on these :/
*Tuesday, October 9, 2012 at 12:11pm*

**Trig!**

just plug and chug where do you get stuck?
*Tuesday, October 9, 2012 at 4:54am*

**Trig!**

Find the following values for each function. A. f(0) B. f(-1) C. f(-x) D. f(x+1) f(x)=2x^2+3x-4 and f(x)=x^2/x+1
*Tuesday, October 9, 2012 at 12:31am*

**math**

I need help with this trig problem cotθ=-√3/3 and 3/2π≤θ≤π, find θ
*Monday, October 8, 2012 at 7:46pm*

**Precalc/Trig Word Problem**

Greg ! Enough of "homework - dumping" Show some work before any more help is given.
*Monday, October 8, 2012 at 9:29am*

**Precalculus/Trig 6**

follow the steps that I showed you in previous post
*Monday, October 8, 2012 at 9:26am*

**Precalculus/Trig 5**

You should be able to make a quick sketch of the following right-angled triangles 30° - 60° - 90° ---> π/6rad - π/3rad - π/2 ---> 1 -- √3 -- 2 and 45° - 45° - 90° --> π/4 - π/4 - π/2 ---> 1 - 1 - √2 ...
*Monday, October 8, 2012 at 9:26am*

**Precalculus/Trig 3**

180° = π radians , (memorize this relationship) so 1° = π/180 radians then 40° = 40(π/180) = ...
*Monday, October 8, 2012 at 9:14am*

**trig**

Make the following diagram Draw a base line , label the balloon as B and above the base line Label the point directly below the balloon A on the base line label the two stations as C and D so that angle ACB = 78.2 and angle ADB = 17.6 Now in triangle BCD , CD = 10 , angle BCD...
*Monday, October 8, 2012 at 9:12am*

**Precalc/Trig Word Problem**

From the top of a 200-ft lighthouse, the angle of depression to a ship in the ocean is 23 degrees. How far is the ship from the base of the lighthouse?
*Monday, October 8, 2012 at 9:12am*

**Precalc/Trig 2**

so you given y = -√3 x^2 + y^2 = 2^2 x^2 + 3 = 4 x^2 = 1 x = ± 1 , but the point is in IV so x = 1
*Monday, October 8, 2012 at 9:05am*

**Precalculus/Trig 6**

Find the exact value of the following trigonometric functions: sin(13pi), cos(14pi), tan(15pi)
*Monday, October 8, 2012 at 9:05am*

**Precalc/Trig 1**

show that (√11/6)^2 + (5/6)^2 = 1 , (it is)
*Monday, October 8, 2012 at 9:03am*

**Precalculus/Trig 5**

Find the exact value of thet following trigonometric functions: tan(5pi/6) tan(7pi/6) tan(11pi/6)
*Monday, October 8, 2012 at 9:03am*

**Precalculus/Trig 3**

The degree measure of an angle theta is 40 degrees. What is the radian Measure of theta? at what point P does the terminal ray intersect the unit circle (what is the ordered paid P(____, ___)?
*Monday, October 8, 2012 at 9:02am*

**trig**

weather balloon is directly west of two observing stations that are 10 mi apart. The angles of elevation of the balloon from the two stations are 17.6 degrees and 78.2 degrees. How high is the balloon?
*Monday, October 8, 2012 at 8:58am*

**Precalc/Trig 2**

The point P lies in the 4th quadrant and is located on a circle of radius 2. Find the missing coordinate for P. (___, (negative square root of 3)).
*Monday, October 8, 2012 at 8:57am*

**Precalc/Trig 1**

What equation would you use to show that the point ((square root of 11)/(6), (5/6))is on the unit circle?
*Monday, October 8, 2012 at 8:53am*

**trig**

let the height be h sin55° = h/170 h = 170sin 55° = ...
*Monday, October 8, 2012 at 8:46am*

**trig**

two intersecting sides are 250ft and 170ft, angle between is 55 degrees. use the 250ft side as the base, what is the height of the triangle
*Monday, October 8, 2012 at 8:28am*

**trig**

(E) none of the above
*Saturday, October 6, 2012 at 7:01am*

**Calculus 2**

just like regular trig functions, sinh^2(x) = (cosh(2x)-1)/2 ∫(cosh(2x)-1)/2 = 1/2 (1/2 sinh(2x) - x)
*Thursday, October 4, 2012 at 2:13pm*

**Calculus 2**

just like regular trig functions, use integration by parts to get x arcsinh(x) - √(x^2+1)
*Thursday, October 4, 2012 at 10:45am*

**trig**

Are we solving for Ø ? tan^2 Ø +6 = sec^2 Ø + 5 sin^2 Ø/cos^2 Ø + 6 = 1/cos^2 Ø + 5 times cos^2 Ø sin^2 Ø + 6cos^2 Ø = 1 + 5 cos^2 Ø sin^2 Ø + cos^2 Ø = 1 1 = 1 ahh, it was an identity, thus ...
*Thursday, October 4, 2012 at 8:36am*

**trig**

tan^2theta+6=sec^2theta+5
*Thursday, October 4, 2012 at 2:34am*

**TRIG**

Thank you
*Tuesday, October 2, 2012 at 5:31pm*

**TRIG**

Ø/360 = 8/20 Ø = 360(8/20) = 144°
*Tuesday, October 2, 2012 at 5:28pm*

**TRIG**

find the degree measure of a central angle subtended by an arc of 8.00cm in a circle with circumference 20.0cm
*Tuesday, October 2, 2012 at 5:22pm*

**trig**

33000 ft = 33000/5280 miles = 5.681818... miles let the angle of descent be Ø tanØ = 5.681818.../130 = .043706... Ø = appr 2.5°
*Monday, October 1, 2012 at 9:11pm*

**Math, Trig**

Thanks so much!
*Monday, October 1, 2012 at 9:07pm*

**Math, Trig**

Draw a "side view" Draw 3 points on the river, B , C, and D , B to the left of the points draw a point A above B so that angle B is 90° Join AC and AC , so that CD = 20 Mark angle ADC = 62.6° and angle ACB = 72.8° label AB =h (h is the height) , BC = x in...
*Monday, October 1, 2012 at 8:46pm*

**trig**

a plane flying 33,000 ft is 130 miles from the airport when it begins to descend if the angle of descent is constant find this angle
*Monday, October 1, 2012 at 8:31pm*

**Math, Trig**

The high level bridge, a railway bridge that crosses the Oldman River is over 1km long. From one point on the river, the angle of elevation of the top of the bridge is 62.6 degrees. From a point 20m closer to the bridge, the angle of elevation of the top of the bridge is 72.8 ...
*Monday, October 1, 2012 at 8:00pm*

**trig**

what's the trouble? Just trace the circle twice. If you have to label some points along the way, just label each marked point with two values: pi/2,5pi/2 and so on
*Friday, September 28, 2012 at 11:24am*

**trig**

i have to draw a flat unit circle all the way to 4 pi in radians. i can get to 2 pi but i have trouble finding the radians if i went one more time around the unit circle.
*Friday, September 28, 2012 at 1:04am*

**Math/Trig**

I am assuming that 0° is east. If you are using 0° = north, then things have to be changed a bit.
*Thursday, September 27, 2012 at 11:36am*

**Math/Trig**

150 at 180° = (-150,0) 170 at 210°50' = (-146,-87) total: (-296,-87) distance = √(296^2 + 87^2) = 309 mi
*Thursday, September 27, 2012 at 11:35am*

**Math/Trig**

An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction 210°50'. Assuming the Earth is flat, how far is the plane from the airport at this time (to the nearest mile)
*Thursday, September 27, 2012 at 10:15am*

**Precalc/Trig**

wave is modeled by the function ... h(t) = 3cos(p/10)*t What is the period of the wave (T) *frequency (F) is defined as the number of cycles of the motion per second. What is the relationship between F and T? Find the wave height (H) which is the vertical distance between the ...
*Wednesday, September 26, 2012 at 9:17am*

**trig**

X = -23 Mi. Y = 14 Mi. tanA = Y/X = 14/-23 = -0.60870 A = -31.33o,CW = 148.7o,CCW. D = X/cosA = -23/cos148.7 = 26.9 Mi Bearing = 148.7 - 90 = 58.7o.
*Tuesday, September 25, 2012 at 7:24pm*

**Trig**

if the ladder has length x, 4/x = cos 40° x = 5.22 ft
*Tuesday, September 25, 2012 at 2:54pm*

**Trig**

A ladder is resting against a wall. The ladder and the ground and angle of 40 degrees and the ladder is 4 ft. from the wall. How long is the ladder?
*Tuesday, September 25, 2012 at 1:53pm*

**science**

Use kinematics to calculate the height for the firstball, which is Vf=0 a=9.8 t=1.5 because you take the total time it take to go up and come back down and divide it by 2. 3/2 = 1.5 getting an initial velocity of Voy= 14.7m/s V = t(9.8) = 1.5(9.8) = 14.7 This must have the ...
*Tuesday, September 25, 2012 at 12:01am*

**Trig**

A sprinkler on a golf green is set to spray water over a distance of 20 meters and to rotate through an angle of 160°. Find the area of the region that can be irrigated with the sprinkler. (Round your answer to two decimal places.) I really have no idea how to go about ...
*Monday, September 24, 2012 at 8:23pm*

**Trig**

Two ships leave the same port at 7.am. The first ship sails towards europe on a 54 degree course at a constant rate of 36 mi/h. The second ship,neither a tropical destination, sails on a 144 degree course at a constant speed of 42 mi/h. Find the distance between the ships at ...
*Monday, September 24, 2012 at 7:01pm*

**trig**

A boat is 23 mi due west of lighthouse A. Lighthouse B is 14 mi due north of lighthousenA. Find the bearing of lighthouse B from the boat and the distance from lighthousenB tho the boat.
*Monday, September 24, 2012 at 6:40pm*

**trig**

A boat is 23 mi due west of lighthouse A. Lighthouse B is 14 mi due north of lighthousenA. Find the bearing of lighthouse B from the boat and the distance from lighthousenB tho the boat.
*Monday, September 24, 2012 at 6:32pm*

**small typo -Trig**

near the 2/3 mark of solution should say so P is (10.8 , 9.6) typo had no effect on rest of solution
*Sunday, September 23, 2012 at 9:36pm*

**Trig**

Put the diagram on the x-y grid so the centre of the circle for the cylinder is C(6,6) Let P(x,y) be the point of contact of the ladder with the cylinder. Slope of ladder touching at P is -3/4 So the slope of PC = 4/3 (y-6)/(x-6) = 4/3 4(y-6) = 3(x-6) y-6 = (3/4)(x-6) #1 ...
*Sunday, September 23, 2012 at 9:34pm*

**Trig**

Draw a picture of a right triangle. The hypotenuse (the longest side opposite of the ninety degree angle) would be 16. The opposite of the desired angle is 12. The sine of what angle equals 12/16 (6/8 3/4)?
*Sunday, September 23, 2012 at 9:09pm*

**Trig**

use the equation y=mx+b
*Sunday, September 23, 2012 at 8:54pm*

**Trig**

A cylindrical tank, 6 foot in radius, lies on it's side parallel and against the side of a warehouse. A ladder leans against the building, passes over and just touches the tank, and has a slope of -3/4. Find the equation of the ladder and the length of the ladder
*Sunday, September 23, 2012 at 8:47pm*

**trig**

A woman that is 5'4" stands 15 ft from a streetlight and casts a four-foot long shadow. Determine the height of the streetlight and the degree measure of the angle of elevation form the tip of her shadow to the top of the streetlight, both accurate to two decimal places.
*Sunday, September 23, 2012 at 5:59pm*

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