Thursday

April 17, 2014

April 17, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**TRIG,**

Can you solve and show me the work in these two equations? cos(3x)=-1 tan (theta)=-1
*Sunday, January 12, 2014 at 10:59pm*

**trig**

write a formula f*g*h when f(x)=x+1, g(x)=3x, h(x)=4-x
*Sunday, January 12, 2014 at 8:07pm*

**Trigonometry**

I did not get this sinx = 2/3, then cosx = √5/3 if secy = 5/3, then cosy = 3/5, and siny = 4/5 sin(x+y) = sinxcosy + cosxsiny = (2/3)(3/5) + (√5/3)(4/5) = (6 + 4√5)/15 Your individual trig values are correct, you must have made a substitution error.
*Friday, January 10, 2014 at 11:59pm*

**trig**

Both answers are correct.
*Friday, January 10, 2014 at 7:06pm*

**Physics Classical Mechanics**

You are welcome. I took trig in the early 1950s but have done a little since :)
*Friday, January 10, 2014 at 6:00pm*

**Physics Classical Mechanics**

.21[ sinT(-sinT)+cos^2T] = .137 sinT cosT .21 (cos^2T - sin^2T) = .137 sinT cosT cos^2T - sin^2T = .651 sinT cosT Oh my, trig identities or trial and error?
*Friday, January 10, 2014 at 5:17pm*

**trig**

d = 50mi * 5280Ft/m1 = 264,000 Ft. = 264K. (0,10k), (264k,20k). Tan A = (20k-10k)/(264k-0) = 0.037879 A = 2.2 Degrees.
*Wednesday, January 8, 2014 at 8:55pm*

**trig**

A pilot is flying at 10,000 feet and wants to take the plane up to 20,000 feet over the next 50 miles. What should be his angle of elevation to the nearest tenth?
*Wednesday, January 8, 2014 at 5:28pm*

**trig**

58585
*Tuesday, January 7, 2014 at 1:27am*

**trig**

After you make your sketch, it should be simple to calculate the angle between their flight-paths to be 88° , and after 2 hours the distance covered would be 650 and 600 miles. So a direct application of the cosine law .... d^2 = 650^2 + 600^2 - 2(650)(600)cos88°...
*Monday, January 6, 2014 at 11:07am*

**trig**

Two Airplanes leave an airport at the same time on different runways. One flies on a bearing of N66degreesW at 325 miles per hour. The other airplane flies on a bearing öf S26degreesW at 300 miles per hour. How far apart will the airplanes be after 2 hours?
*Monday, January 6, 2014 at 9:59am*

**algebra 2 trig**

log 300 = log 3 + log 100 = a+2
*Saturday, January 4, 2014 at 5:22pm*

**algebra 2 trig**

if log3=a then log300 can be expressed as? 1) 100a 2)a+2 3)100+a 4)3a
*Saturday, January 4, 2014 at 4:54pm*

**trig**

I am answering this from a Starbucks on my tablet and have no access to a calculator to check your answer.
*Thursday, January 2, 2014 at 10:14am*

**trig**

I assume that the bearing is from the position of the ship. Then the angle opposite the 7 side is 18 degrees. Sin@/12 = sin18/7 Gives you the bottom angle
*Thursday, January 2, 2014 at 10:08am*

**trig**

the triangle show with sides 12 and 7 and co-contained angle of 62 degrees adjascent to 12 sides by sin law: sin@/12=sin62/7 is this correct
*Thursday, January 2, 2014 at 8:51am*

**trig**

so the second angle is 17.46 and the third angle is 133.54. then it travels to NortEast direction, am I correct?
*Thursday, January 2, 2014 at 8:33am*

**trig**

so the second angle is 17.46 and the third angle is 133.54. then it travels to NortEast direction, am I correct?
*Thursday, January 2, 2014 at 8:29am*

**trig**

I have helped you with 3 out of your 4 questions, they are all similar to each other. Give this one a try and let me know what you think.
*Thursday, January 2, 2014 at 8:09am*

**trig**

40° 5' = (40 + 5/60) = 481/12 ° You will have two right-angled triangle For the distance of the closer object tan (481/12)° = 57/d1 d1 = 57/tan(481/12)° For the distance of the farther away object tan 27 = 57/d2 d2 = 57/tan27° distance between objects...
*Thursday, January 2, 2014 at 8:08am*

**trig**

After the 10 mile sailing, let the distance between boat and lighthouse be x by the cosine law, you can find x x^2 = 14^2+10^2 - 2(14)(10)cos57° , (90°-33°=57°) Once you have x, you can find a second angle using the sine law, and voila!, easy after that.
*Thursday, January 2, 2014 at 8:01am*

**trig**

My diagram shows a triangle with sides 75 and 120 and a con-contained angle of 29° adjacent to the 75 side by sine law: sinØ/75 = sin29°/120 sinØ = 75sin29°/120 Once you have the second angle Ø , it is easy to find the third angle and answer ...
*Thursday, January 2, 2014 at 7:53am*

**trig**

From a tower 57 ft high two objects in a straight line from it are sighted at angles of depression of 27degrees and 40degrees5' respectively. Find the distance between the two objects.
*Thursday, January 2, 2014 at 7:40am*

**trig**

A lighthouse is 14 miles east of a dock. A ship sails N33degreesE from the dock. What will be its bearing from the lighthouse after having sailed 10 miles?
*Thursday, January 2, 2014 at 7:33am*

**trig**

Illustrate and solve. A monument near a dock is 12 miles east of a ship. After the ship has sailed 7 miles, the monument bears N62degreesE. In what direction is the ship sailing?
*Thursday, January 2, 2014 at 7:26am*

**trig**

A plane flies 75 miles south from an airfield, and then travels 120 miles more in a different direction. By this time it bears S29degreesE of the airfield. In what direction is it heading. illustrate and solve.
*Thursday, January 2, 2014 at 7:19am*

**trig**

I assume you made a sketch. On mine I let the distance between tower and cliff be x m then tan60° = 100/x x = 100/tan60° In my smaller right-angled triangle at the top of my diagram , let the difference between height of tower and height of cliff be h then tan 30 = h/x...
*Saturday, December 28, 2013 at 10:31am*

**trig**

The angle of depression of the top and bottom of a tower as seen from the top of a 100m high cliff are 30degree and 60degree respectively. find the height of the tower
*Saturday, December 28, 2013 at 9:58am*

**3-D Coordinates Polar coordinates Precalculus Trig**

Sorry, never mind. I figured it out.
*Tuesday, December 24, 2013 at 10:22am*

**3-D Coordinates Polar coordinates Precalculus Trig**

Hi. I don't understand why you multiply by 50*pi/180. The 50 degrees is the difference in the longitude, but why isn't that divided by 360? Why the pi/180? If someone could clear that up I would appreciate it.
*Tuesday, December 24, 2013 at 10:18am*

**3-D Coordinates Polar coordinates Precalculus Trig**

thank you that helped !
*Wednesday, December 18, 2013 at 1:00pm*

**3-D Coordinates Polar coordinates Precalculus Trig**

Along the equator, if you travel an angle of θ, your distance is rθ. The travel is along the 40° latitude, so moving through an angle of θ° of longitude, you only travel rθcos40°. The Lollipop moved 50°, so it traveled (6400)(50*π/180)(...
*Wednesday, December 18, 2013 at 11:23am*

**3-D Coordinates Polar coordinates Precalculus Trig**

The U.S.S. Lollipop is on assignment in the Atlantic Ocean. It travels from a longitude of 70 degrees west to 20 degrees west, along the latitude of 40 degrees north. How far does it travel? (Assume that the radius of the Earth is 6,400 km.) Erm I don't know how to solve ...
*Wednesday, December 18, 2013 at 10:59am*

**Please help trig proof**

LS = (cotx + cscx)/(sinx + tanx) = (cosx/sinx + 1/sinx)/(sinx + sinx/cosx) = ( (cox + 1)/sinx) / ( (sinxcosx + sinx)/cosx) = (cosx + 1)/sinx * cox/(sinx(cosx + 1) = cotx (cosx + 1)/(sinx(cosx + 1) ) = cotx (1/sinx) = cotx cscx = RS
*Monday, December 16, 2013 at 9:31pm*

**Please help trig proof**

((cotx+cscx)/(sinx+tanx))=cotxcscx Please prove left side equal to right side, only doing doing work on the left.
*Monday, December 16, 2013 at 8:55pm*

**Simplifying trig expression**

You could actually prove this for yourself by using one of the expansion formulas recall sin(A+B) = sinAcosB + cosAsinB so sin(x+90) = sinx cos90 + cosx sin 90 = sinx (0) + cosx (1) = cos x test it with some angles you should know let x = 30° sin (30+90) = sin 120 = √...
*Monday, December 16, 2013 at 2:54pm*

**Simplifying trig expression**

sin(x+90) = ? cos(x+90) = ? Would the first equal like -sin(x)? Would the second equal uh -cos(x)? I'm not quite sure :/
*Monday, December 16, 2013 at 2:36pm*

**Trigonometry**

draw a horizontal line from the top of the tree to the building to create 2 right-angled triangle That line is common to both So, set up a trig equation for each triangle using that common side. You can then use substitution to get your answer.
*Sunday, December 15, 2013 at 10:45am*

**Math - Trig**

a) after you make your sketch you should see that cos A = 120/d b) if d=1500 cosA = 120/1500 = .08 A = appr 5.4° ( here is what I did on my calculator ..... 120 ÷ 1500 = 2ndF cos = to get 85.4114... c) repeat the steps I showed you in b) let me know what you got
*Saturday, December 14, 2013 at 9:01pm*

**Math - Trig**

Assume the observations were made from the same building. Let the distance of the building from the tree is D in m. Then D(tan22+tan50)=150 D=(150)/(tan22+tan50)
*Saturday, December 14, 2013 at 8:02pm*

**Math - Trig**

From the foot of a building I have to look upwards at an angle of 22degrees to sight the top of a tree. From the top of a building, 150 meters above ground level, I have to look down at an angle of depression of 50degrees to look at the top of the tree. a. How tall is the tree...
*Saturday, December 14, 2013 at 7:36pm*

**Math - Trig**

An observer on the ground at point A watches a rocket ascend. The observer is 120 feet from the launch point B. As the rocket rises, the distance d from the observer to the rocket increases. a. Express the measure of angle A in terms of d. b. Find the measure of angle A if d...
*Saturday, December 14, 2013 at 7:34pm*

**trig**

let AP be the vector of magnitude 23 lbs and direction 18° and BP be the vector of magnitude 27 lbs and direction -15° Complete the parallogram APBQ , with AP=2 BP=27 angle P = 33° then angle B = 180-33 = 147° So PQ is the force pulling by the puppy to hold the...
*Thursday, December 12, 2013 at 10:01pm*

**trig**

Juana and Diego Gonzales, ages six and four respectively, own a strong and stubborn puppy named Corporal. It is so hard to take Corporal for a walk that they devise a scheme to use two leashes. If Juana pulls with a force of 23 lbs at an angle of 18Â° and Diego ...
*Thursday, December 12, 2013 at 9:51pm*

**math - trig**

The angle of elevation is 350 ??? I suspect a typo. Did you mean 35° ? Correct your typo, then set it up like I showed you in the previous post.
*Thursday, December 12, 2013 at 8:11am*

**math - trig**

Always make a sketch. draw a horizontal showing 250 ft below the water line mark it as 1500 draw a vertical at the end of the horizontal to show the depth of the dive, label it x label the angle of the dive as 15° Now, in relation to the Ø, isn't the horizontal ...
*Thursday, December 12, 2013 at 8:09am*

**math - trig**

tanØ = 2059/5280 = ... Ø = 21.3°
*Thursday, December 12, 2013 at 7:59am*

**math - trig**

The airport meteorologists keep an eye on the weather to ensure the safety of the flights. One thing they watch is the cloud ceiling. The cloud ceiling is the lowest altitude at which solid cloud is visible. If the cloud ceiling is too low the planes are not allowed to take ...
*Thursday, December 12, 2013 at 6:06am*

**math - trig**

A submersible traveling at a depth of 250 feet dives at an angle of 15º with respect to a line parallel to the water’s surface. It travels a horizontal distance of 1500 feet during the dive. What is the depth of the submersible after the dive?
*Thursday, December 12, 2013 at 6:00am*

**math - trig**

The tallest television transmitting tower in the world is in North Dakota, and it is 2059 feet tall. If you are on level ground exactly 5280 feet (one mile) from the base of the tower, what is your angle of elevation looking up at the top of the tower?
*Thursday, December 12, 2013 at 6:00am*

**Math - Trig**

assuming all these buildings are the same one, if the building is at distance d from the tree of height h, draw a diagram to see that (150-h)/d = tan 50° h/d = tan 22° plugging in the numbers, h+1.19d = 150 h = 0.40d So, we have h = 94.3 d = 37.7
*Thursday, December 12, 2013 at 5:11am*

**Math - Trig**

From the foot of a building I have to look upwards at an angle of 22° to sight the top of a tree. From the top of a building, 150 meters above ground level, I have to look down at an angle of depression of 50° to look at the top of the tree. a. How tall is the tree? b...
*Thursday, December 12, 2013 at 3:20am*

**Math**

standard trig substitution Just as ∫[dy/(1+y^2)] = arctan(y), ∫[dy/(1-y^2)] = arctanh(y)
*Tuesday, December 10, 2013 at 11:40pm*

**Tan^2 Integrals**

don't forget your trig identities tan^2+1 = sec^2 ∫sec^2 y dy is easy, right? 7tan^2 u + 15 = 7tan^2 u+7 + 8 = 7sec^2 u + 8 Now we're back to ∫7sec^2 u + 8 du
*Tuesday, December 10, 2013 at 11:05am*

**GEOMETRY CIRCLES**

#2 Notice the centre of the inscribed circle must lie on the bisectors of the angles. We have an isosceles triangle, with sides 10,10 and 12 If we place the triangle on the x-y grid, we can have the base from (-6,0) to (6,0) and the third vertex on the y-axis. The y-axis ...
*Tuesday, December 10, 2013 at 10:27am*

**Math: Trig**

Looks good to me.
*Tuesday, December 10, 2013 at 6:08am*

**Math: Trig**

I figured it out and think both have to be positive so the answer is D. Let me know if I'm wrong
*Tuesday, December 10, 2013 at 2:00am*

**Math: Trig**

Thanks. For #3, I did not realize he gave me multiple choice answers. A) (-13.25, 21.20) B) (-21.20, -13.25) C) (13.25, 21.20) D) (21.20, 13.25)
*Tuesday, December 10, 2013 at 12:55am*

**Math: Trig**

(x,y) = (2cos 4π/11,2sin 4π/11) P = (√26,arctan(-1/-5)) v = 25cos32°i + 25cos32°j u+v = <(6-10),(2+1)> -3u = <-3(6),-3(2)>
*Tuesday, December 10, 2013 at 12:37am*

**Math: Trig**

Help with vectors, please. Thanks 1) Express point in rectangular form. Give EXACT answer, if possible? (2, 4π /11) 2) Given P = (2. -3) and Q = (-3, -4). find the component form of vector PQ 3) The vector v has a magnitude of 25 inches and a direction of 32° , ...
*Tuesday, December 10, 2013 at 12:25am*

**Trig**

Hard to say, due to the typos, but it looks like you want tan(105-(-15)) = tan(120) = -tan(60) = -√3
*Monday, December 9, 2013 at 11:23pm*

**Trig**

86.16 i believe
*Monday, December 9, 2013 at 9:11pm*

**Trig**

Sum and diffence formula Finding Exact value of Tan 105-Tan 10)-15)/1+ tan(105)Tan(-15)
*Monday, December 9, 2013 at 8:59pm*

**trig**

take a look here: http://math.ucsd.edu/~wgarner/math4c/textbook/chapter6/product_sum_formulas.htm
*Monday, December 9, 2013 at 12:00am*

**Trigonometry**

The reason they come up in trig is that any point in the plane can be located at some distance from the origin, and in some direction, given by an angle θ. All your normal trig functions can be applied to θ.
*Sunday, December 8, 2013 at 11:45pm*

**trig**

prove that: sinC+sinD =2sin((C+D)/2)*cos((C-D)/2)
*Sunday, December 8, 2013 at 9:49pm*

**Trig**

if the 60km is horizontal distance, then tan x = 10/60 if it's line-of-sight distance, then sin x = 10/60 luckily, for small x, sinx and tanx are very close to each other.
*Saturday, December 7, 2013 at 11:08pm*

**trig**

actually i did it already
*Saturday, December 7, 2013 at 10:27pm*

**trig**

Points X, Y, and Z are on the circumference of a circle with radius 2 such that <YXZ = 45 degrees and <XZY = 60 degrees. Find the area of triangle XYZ.
*Saturday, December 7, 2013 at 9:08pm*

**Trig**

To avoid a steep descent, an airplane flying at 10,000m starts its descent 60km away from the airport. For the angle of descent to be constant, at what angle should the plane descend?
*Saturday, December 7, 2013 at 8:08pm*

**Trigonometry/Geometry - Inequalities**

Let a, b, and c be positive real numbers. Prove that sqrt(a^2 - ab + b^2) + sqrt(a^2 - ac + c^2) is greater or equal to sqrt(b^2 + bc + c^2). Under what conditions does equality occur? That is, for what values of a, b, and c are the two sides equal? This looks like a geometry...
*Saturday, December 7, 2013 at 4:58pm*

**Trig-Weird Geometry Problem**

Thanks steve!
*Friday, December 6, 2013 at 11:14am*

**Math-Trig**

Thanks much :)
*Thursday, December 5, 2013 at 8:52pm*

**Math-Trig**

1. The first thing you need to do is convert 60km to meters, or 10,000m to kilometers; whichever you prefer. Next,it helps if you draw a right triangle. The plane is at a height of 10,000m; this is the vertical leg of the triangle. It is 60 km away from the airport; this is ...
*Thursday, December 5, 2013 at 8:41pm*

**Math-Trig**

1. To avoid a steep descent, an airplane flying at 10 000 m starts its descent 60km away from the airport. For the angle of descent to be constant, at what angle should the plane descend? 2. A flagpole that is 20m high casts a shadow that is 18m long. What is the angle of ...
*Thursday, December 5, 2013 at 8:09pm*

**Math - Trig**

50/x = tan 63 x/20000 = sin 15.33
*Thursday, December 5, 2013 at 7:38pm*

**Math - Trig**

a, Tan 27=50/d solve for d b. sinAngle=h/20,000 solve for h.
*Thursday, December 5, 2013 at 7:37pm*

**Math - Trig**

1. When you look down from the top of a building at an angle of 63degrees, you will see a man reading a newspaper. If the building is 50m high, how far is the man from the building? 2. A plane takes off at an angle of 15deg20'. How high will it have risen after it has ...
*Thursday, December 5, 2013 at 7:31pm*

**Trig-Weird Geometry Problem**

You will find a nice discussion at http://answers.tutorvista.com/1101359/trigonometry-application-rule.html showing that if A+B+C=180, tan(A-B)+tan(B-C)+tan(C-A) = tan(A-B)*tan(B-C)*tan(C-A) Given that, one of the factors must be zero. So, A=B or B=C or A=C. That is, ...
*Thursday, December 5, 2013 at 12:44pm*

**Trig-Weird Geometry Problem**

Determine all triangles ABC for which tan(A-B)+tan(B-C)+tan(C-A)=0. There's a hint: "Can you relate A-B to B-C and C-A?" Should I apply the tangent difference formula (tan(x-y))? Help would be appreciated, thanks.
*Thursday, December 5, 2013 at 11:02am*

**Advanced Functions/Precalculus**

Trigonometry Questions 1.) Find the exaqct value of tan(11π/12) 2.) A linear trig equation involving cosx has a solution of π/6. Name three other possible solutions 3) Solve 10cosx=-7 where 0≤x≤2π
*Wednesday, December 4, 2013 at 2:13pm*

**Math-Trig**

make a diagram showing a sideview. remember the angle of depression from the helicopter view is equal to the angle of elevation from groundview. So I have a right-angled triangle, with a height of 105 (opposite side), the adjacent side of x and an angle of 30.5 take it from there
*Wednesday, December 4, 2013 at 8:42am*

**Math-Trig**

A helicopter hovers 105 m above the end of an island. If the angle from the helicopter down to the other end of the esland is 13deg30', find the length of the island.
*Wednesday, December 4, 2013 at 8:19am*

**Math - Trig**

Looks like you need to review the basic trig functions and draw useful diagrams. #1 h/15 = tan 46.48° #2 h/17.2 = tan 73.5°
*Wednesday, December 4, 2013 at 6:19am*

**Math - Trig**

1. From a point 15m from the base of a tree, the angle of elevation of the top of the tree is 46.48degrees. Approximate the height of the tree. 2. From a point 17.2m from the base of a building, the angle of elevation of the top of the building is 73.5degrees. Aproximate the ...
*Wednesday, December 4, 2013 at 12:11am*

**Trigonometry**

You have received many replies to your numerous trig questions, and have not responded to a single one of them. How do we know that you are even reading the responses, and if so, if they are of any help to you. What is your main difficulty with these questions? What are you ...
*Thursday, November 28, 2013 at 10:59pm*

**Trigonometry functions**

David, really now ! After getting help for so many trig question, don't you think it is time that you try some of these yourself? Especially the rather simple ones, like your last two? Let us know what you get and we'll check them
*Thursday, November 28, 2013 at 9:36pm*

**trig**

if cot 2θ = 5/12, the 2θ is in QI or QIII. But, since 2θ<π, we must be in QI. so, sin2θ = 12/13, cos2θ = 12/13 Now use the half-angle formulas to find sinθ, cosθ, tanθ.
*Thursday, November 28, 2013 at 7:42am*

**trig**

find cos(θ)ˏsin(θ)ˏtan(θ), if cot (2θ)=5/12 with 0≤2θ≤π Answer this Que
*Wednesday, November 27, 2013 at 11:57pm*

**trig.**

The tree top is 100*tan(52) feet higher than the person's eyes. To get the height h, above ground, h=100tan(52)+5 feet = 133 feet, approximately
*Wednesday, November 27, 2013 at 5:55pm*

**trig.**

a person standing 100ft. from the base of the tree looks up to the top of the tree with an angle of elevation of 52. assuming that the persons eyes are 5ft above ground the ground how tall is the tree?
*Wednesday, November 27, 2013 at 4:31pm*

**Math - Trig**

since the altitude divides the base in half, and the vertex angle in half, h/92.1 = cot 21°55'
*Wednesday, November 27, 2013 at 6:09am*

**Math - Trig**

RS/sin T = RT/sin S now just plug in your numbers.
*Wednesday, November 27, 2013 at 6:07am*

**Math - Trig**

1. Given that points S and R on opposite sides of a lake, triangle SRT is formed. To find the distance RS across the lake, a surveyor lays off RT = 53.1 m, with angle T = 32deg 10' and anagle S = 57deg 50'. Find length RS.
*Wednesday, November 27, 2013 at 2:01am*

**Math - Trig**

1. Find the altitude of an isocles triangle having base 184.2 cm if the angle opposite the base is 68deg 44' 2. A 13.5m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the top of the fire truck) if the ladder makes an ...
*Wednesday, November 27, 2013 at 1:58am*

**Trigonometry**

Do you not have a calculator? if cotØ = .791 then tanØ = 1/.791 = appr 1.264 Ø = 51.656° secØ = 1/cosØ = 1.612 2. sin^2 Ø + cos^2 Ø = 1 , one of the major trig identities, which you MUST know.
*Tuesday, November 26, 2013 at 11:48pm*

**Trig-Geometry - Law of sines and cosines**

Hello everyone, I've been struggling on this problem for quite some time. It would be appreciated if you could help. Thanks. (The website is the diagram, it is a screenshot) In the diagram below, triangle ABC has been reflected over its median AM to produce triangle AB'...
*Sunday, November 24, 2013 at 11:01pm*

**Trigonometry**

if you don't understand it at all, you seriously need to review the most basic trig identities: sin^2 + cos^2 = 1 tan = sin/cos cot = 1/tan With that in mind, and suppressing all the x's for ease of reading, (a) sin*tan = sin/cot = sin*cot/cot^2 = sin(cos/sin) / cot^2...
*Sunday, November 24, 2013 at 6:55pm*

**Trig/Precalc**

the ratio of height to shadow length is the same, so h/3.5 = 8/5 I imagine you can solve that for h.
*Sunday, November 24, 2013 at 6:47pm*

**Trig/Precalc**

a lamp post that is 8 feet high casts a shadow 5 feet long. how tall is the person standing beside the lamp post if his shadow is 3.5 feet long
*Sunday, November 24, 2013 at 6:06pm*

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