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April 20, 2014

April 20, 2014

**Recent Homework Questions About Trigonometry**

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**trig**

The answer is 154 ft
*Tuesday, March 18, 2014 at 1:42pm*

**trigonometry**

oops. Forgot all those hours θ = 2π*5 + 2π/3 = 10π + 2π/3 = 32π/3 For the minute hand, that's the same value as 2π/3 but, θ/12 is now 32π/36 = 8π/9 which affects its trig function values immensely.
*Tuesday, March 18, 2014 at 6:27am*

**trig**

2tanA+6=tanA+3 tanA = -3 so A must be in II or IV using tanx = +3 x = 71.565° so A = 180-71.564 = appr 108.4 or A = 360 - 71.565 = appr 288.4°
*Sunday, March 16, 2014 at 9:54pm*

**trig**

Find all angles, 0≤A<360, that satisfy the equation below, to the nearest 10th of a degree. 2tanA+6=tanA+3
*Sunday, March 16, 2014 at 9:16pm*

**Math ( trig )**

bottom of wheel up 5 so height above wheel bottom = 40 distance below center = 47 -40 cos A = 7/47 A = 81.4 degrees
*Wednesday, March 12, 2014 at 5:12pm*

**Math ( trig )**

If a person on the Ferris wheel is 45 feet above the ground. The Gris wheel has a radius of 47.f and is 5 feet above the ground. At what degrees had the Ferris wheel rotated counted clockwise?
*Wednesday, March 12, 2014 at 3:59pm*

**trig**

mm
*Monday, March 10, 2014 at 3:10pm*

**math trig**

If B is at distance x from the tower of height h, h/x = tan 46° h/(x+80) = tan21° so, eliminating x, we get h/tan46° = h/tan21° - 80 Now just solve for h
*Monday, March 10, 2014 at 5:37am*

**trig**

h/80 = sin 40°
*Monday, March 10, 2014 at 4:59am*

**trig**

a kite is flying at an angle of elevation of about 40 degrees. all 80 meters of string have been let out. ignoring the sag in the string, find the height of the kite to the nearest 10 meter
*Sunday, March 9, 2014 at 11:15pm*

**math trig**

draw the figure. let h height of momnument. Let d be the distance from the bottom of the tower to the bottom of the monument. Draw a dotted horizonal plane from the top of the monument to the tower. Let the distance from this line to the top of the tower be 145-h from the ...
*Sunday, March 9, 2014 at 8:13pm*

**math trig**

A TOWER AND A MONUMENT STAND ON A LEVEL PLANE.THE ANGLES OF DEPRESSION OF THE TOP AND BOTTOM OF THE MONUMENT VIEWED FROM THE TOP OF THE TOWER ARE 13DEGREE AND 31 DEGREE,RESPECTIVELY;THE HEIGHT OF THE TOWER IS 145 FT. FIND THE HEIGHT OF THE MONUMENT.
*Sunday, March 9, 2014 at 7:43pm*

**math trig**

two points A and Bare 80 feetapart on the same side of A towerand on a horizontal line through its foot.if angle of elevation of the top of the tower at A is 21 degree and B is 46 degree find the height of the tower.
*Sunday, March 9, 2014 at 7:40pm*

**trig**

one resultant
*Sunday, March 9, 2014 at 6:53pm*

**trig**

three forces
*Sunday, March 9, 2014 at 4:46pm*

**Trig**

no. sinA = 2 sinB
*Friday, March 7, 2014 at 12:36pm*

**Trig**

In my sketch I see a non-rightangled triangle with angles 42° , 110° and 28° with the side opposite the 42° as the guy wire by the sine law: x/sin42 = 25/sin28 x = 25sin42/sin28 = .....
*Friday, March 7, 2014 at 8:27am*

**Trig**

A guy wire attached to the top of an electric pole makes a 70deg angle with the level ground. At a point 25 feet from the guy wire (farther away from the pole), the angel of elevation to the top of the pole is 42deg. How long is the guy wire?
*Friday, March 7, 2014 at 4:34am*

**Trig**

absurd angles, check your typing. did you mean 70° instead of 700 ?
*Friday, March 7, 2014 at 1:01am*

**Trig**

Given ∆ABC. If side a is twice as long as side b, is <A necessarily twice as large as <B? Why?
*Thursday, March 6, 2014 at 10:27pm*

**Trig**

A guy wire attached to the top of an electric pole makes a 700 angle with the level ground. At a point 25 feet from the guy wire (farther away from the pole), the angel of elevation to the top of the pole is 420. How long is the guy wire?
*Thursday, March 6, 2014 at 10:26pm*

**trig**

city a is 300 miles directly north of city b assuming the earth to be a sphere of radius 4000 miles determine the difference in latitude of the two cities make answers accurate to the nearest second
*Wednesday, March 5, 2014 at 8:21pm*

**Trig**

ok.... so what is the answer?
*Wednesday, March 5, 2014 at 1:07pm*

**Trig**

tan 37 = h/x tan 58 = h/(5-x) x tan 37 = (5-x) tan 58 x (tan 37+tan 58) = 5 tan 58 x = 5 tan 58/ (tan 58 + tan 37) then h = x tan 37
*Wednesday, March 5, 2014 at 12:55pm*

**Trig**

A weather balloon is sighted between points A and B which are 5 miles apart on level ground. The angle of elevation of the balloon from A is 37 degrees and it's angle of elevation from B is 58 degrees. Find the height, in feet, of the balloon above the ground.
*Wednesday, March 5, 2014 at 12:52pm*

**Trig**

2 sin 3x = 2[ 3 sin x-4 sin^3 x] = 6 sin x - 8 sin^3 x 4 sin 2x = 4[ 2 sin x cos x ] = 8 sin x cos x so 6 sin x - 8 sin^3 x + 8 sin x cos x 2 sin x [ 3 - 4 sin^2 x + 4 cos x ]
*Wednesday, March 5, 2014 at 11:19am*

**Trig**

How do you solve this problem? y = 2sin 3x + 4sin 2x I haven't done these types of problems for awhile and have forgotten how to do them.
*Wednesday, March 5, 2014 at 7:52am*

**Math**

oh, please. What is the first and most fundamental trig identity you ever learned? cos^2 + sin^2 = 1 right? Time for the duhh! and head slap.
*Monday, March 3, 2014 at 7:14pm*

**Geometry/Trig**

assuming the crosswind is at right angles to the flight, √(130^2+20^2) = ...
*Monday, March 3, 2014 at 12:26am*

**Geometry/Trig**

A plane is flying and show an air speed of 130 mi/h. However, there is a 20 mi/hr crosswind. What is the resulting speed of the plane.
*Monday, March 3, 2014 at 12:24am*

**Trig**

sorry typo cos x cos y + sin x sin y 5 sqrt 7/52 + 9/13 = .9467
*Sunday, March 2, 2014 at 7:38pm*

**Trig**

cos (x-y) = cos x cos y = sin x sin y cos x = -sqrt7 /4 cos y = -5/13 sin x = 3/4 sin y = 12/13 so cos(x-y)=5sqrt7/52 - 9/13
*Sunday, March 2, 2014 at 7:12pm*

**Trig**

given that sinx=3/4 and cosy=-5/13 and both x and y are in quadrant II, find the exact value of cos(x-y)
*Sunday, March 2, 2014 at 5:54pm*

**Trig**

Thanks to the both of you :)
*Saturday, March 1, 2014 at 8:46am*

**Trig**

or, if you divide top and bottom by cosØ you have (secØ-tanØ)/(secØ+tanØ) now multiply top and bottom by (secØ-tanØ) and you have (secØ-tanØ)^2 / (sec^2Ø-tan^2Ø) = sec^2Ø - 2secØ...
*Saturday, March 1, 2014 at 6:09am*

**Trig**

That's a better job of typing it. I did the first of these in your previous post.. the 2nd: LS = (1-sinØ)/(1+ sinØ) = (1-sinØ)/(1+ sinØ) * (1-sinØ)/(1- sinØ) = (1 - 2sinØ + sin^2 Ø)/(1 - sin^2 Ø) = (1 - 2sin&...
*Friday, February 28, 2014 at 10:06pm*

**Trig**

Prove the following: 1/(tanØ - secØ ) + 1/(tanØ + secØ) = -2tanØ (1 - sinØ)/(1 + sinØ) = sec^2Ø - 2secØtanØ + tan^2Ø
*Friday, February 28, 2014 at 9:41pm*

**Trig**

Don't try to build up fractional equations using your method, the spacing will not work out instead do something like this: 1/(tanØ - secØ) + 1/(tanØ + secØ) = -2tanØ LS = (tanØ + secØ + tanØ - secØ)/(tan2 &...
*Friday, February 28, 2014 at 9:13pm*

**Trig**

Verify the given equations: ____1______ + ____1_______ = -2 tanθ tanθ – secθ tanθ + secθ 1 – sinθ = sec2θ – 2 secθ tan θ + tan2θ 1 + sinθ
*Friday, February 28, 2014 at 8:53pm*

**Geometry**

If all we have is one side and one angle, there's no way to determine any of the missing info. Now, if we're working with a right triangle, then the side opposite the angle is x/59 = tan 35° But since this is geometry, not trig, I don't know whether that's ...
*Thursday, February 27, 2014 at 11:53pm*

**Math (Trig.)**

32
*Thursday, February 27, 2014 at 5:54pm*

**trig**

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, February 27, 2014 at 1:55pm*

**trig**

From the foot of a building i have to look upwards at an angle of 22 degrees 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 degrees to look at the top of the tree. A) How tall is the ...
*Thursday, February 27, 2014 at 1:52pm*

**Alg/Trig**

well, you know that h/x = tan 39.8° h/(x+51.7) = tan 19.7° Now eliminate x and solve for h: h/tan39.8° = h/tan19.7° - 51.7 1.2h = 2.8h - 51.7 h = 32.3 check my math, but that's the method to my madness.
*Wednesday, February 26, 2014 at 12:06pm*

**Alg/Trig**

The angle of elevation from a point on the ground to the top of a tree is 39.8 degrees. The angle of elevation from a point 51.7 ft farther back to the top of the tree is 19.7 degrees. Help find the height of the tree.
*Wednesday, February 26, 2014 at 11:17am*

**trig**

tan 31 = h/580
*Wednesday, February 26, 2014 at 10:39am*

**trig**

a sailor on a boat 580 metres from the base of a vertical cliff sights the top of a lighthouse with an angel of elevation of 31 degrees
*Wednesday, February 26, 2014 at 8:17am*

**trig**

an=(-1)n+1/2n
*Tuesday, February 25, 2014 at 11:21pm*

**trigonometry**

better just stick with "trig" One of the first things to do is to learn the special angles with easy trig values: 0,π/6,π/4,π/3,π/2 Knowing those, you will see that sin π/3 = √3/2 You have -√3/2, so that means that your angle is ...
*Monday, February 24, 2014 at 11:30pm*

**trig**

Hmmm. then sinTheta =1/1.7516=0.571 now try your calculator arcsin, or sin^-1 funcition. make certain you are in degrees mode.
*Sunday, February 23, 2014 at 7:01pm*

**trig**

Use a calculator to find: θ if θ = csc-1 1.7516. θ = _____°. Round to three decimal places.
*Sunday, February 23, 2014 at 5:42pm*

**math12A TRIG pleas help**

since the hypotenuse is 200, the legs are 200 sin 43.2° and 200 cos 43.2° calculate those and just add up the 3 sides.
*Thursday, February 20, 2014 at 5:40am*

**math12A TRIG pleas help**

a farmer wishes to fence a field in the form of a right triangle.If one angle of the right triangle is 43.2 degree and the hypotenuse is 200yard,find the amount of fencing needed.
*Thursday, February 20, 2014 at 2:10am*

**Trig**

E arctan(49/48) S = E 45°35' S = 135°45' Actually, that is the pilot's heading. It also happens to be the bearing of the airport from the plane.
*Wednesday, February 19, 2014 at 12:47pm*

**Trig**

A plane is 48 miles west and 49 miles north of an airport. The pilot wants to fly directly to the airport. What bearing should the pilot take? In degrees and minutes
*Wednesday, February 19, 2014 at 12:09pm*

**Trig**

draw a diagram. Label these points: T = top of pole B = base of pole on road S = tip of shadow on road Draw a horizontal line from S to where it intersects the extension of TB below the road. Call that point C. Now, we want the height h = BT Let x = SC y = CB Now, we have y/60...
*Monday, February 17, 2014 at 4:58am*

**Trig**

A straight road slopes upward 14 degrees from the horizontal. A vertical telephone pole beside the road casts a shadow of 60 feet down the road. if the angle of elevation of the sun is 55 degrees, what is the height of the telephone pole?
*Monday, February 17, 2014 at 4:30am*

**Trig**

well, that would be cos^2(pi/15) + sin^2(pi/15) = 1 In fact, (f+g)(x) = 1 for any x.
*Saturday, February 15, 2014 at 11:12pm*

**Trig**

If f(x)=cos^2x and g(x)=sin^2x, what is (f+g)(pi/15)
*Saturday, February 15, 2014 at 7:29pm*

**trig**

If $5000 is invested at a rate of 3% interest compounded quarterly, what is the value of the investment in 5 years?
*Wednesday, February 12, 2014 at 8:10pm*

**trig**

since sin>0 and cos<0, we are in QII. So, y=1,r=4,x=-√15 sin = y/r = 1/4 cos = x/r = -√15/4 tan = y/x = -1/√15 ...
*Tuesday, February 11, 2014 at 11:24pm*

**trig**

Given csc theta= 4, cot theta < 0 Find the exact values of cos theta and tan theta
*Tuesday, February 11, 2014 at 9:35pm*

**Calculus (Math)**

ah! good one! Looks easy, but there's some algebra and trig identities involved. Rather than slog through it here, I refer you to http://www.wolframalpha.com/input/?i=integral+%28cot4x%29^5+%28csc4x%29^7 and hit the "step by step solution" button.
*Tuesday, February 11, 2014 at 12:23am*

**math - trig**

sin(x) and sin(180-x) have the same value. gotta use the law of cosines, which takes into account the change of sign for obtuse angles.
*Monday, February 10, 2014 at 8:40pm*

**math - trig**

Why cant we solve an oblique triangle with the Law of Sines if we are given SAS?
*Monday, February 10, 2014 at 8:34pm*

**trig**

Note that angle ABC = 33 degrees. So, using the law of cosines, your distance d is given by d^2 = 300^2 + 200^2 - 2(300)(200) cos33 plug and chug
*Monday, February 10, 2014 at 2:03pm*

**trig**

City A is 300kilometer due east of city B. city C is 200kilometer on a bearing of 123degree from city B.how far is it from C to A
*Monday, February 10, 2014 at 1:59pm*

**Math Urgent help please**

As always, draw a diagram. Now, recalling your trig functions, it is easy to see that h/(15+5) = tan 50° now, just solve for h
*Monday, February 10, 2014 at 12:42pm*

**trig**

h = 80*sin40
*Friday, February 7, 2014 at 8:12pm*

**trig**

You are welcome :)
*Thursday, February 6, 2014 at 3:26pm*

**trig**

thank you
*Thursday, February 6, 2014 at 3:23pm*

**trig**

d = r theta = (d/2)theta theta = 2 radians
*Thursday, February 6, 2014 at 3:21pm*

**trig**

a central angle intercepts an arc on a circle equal in length to the diameter of the circle. find the measure, in radians, of the central angle.
*Thursday, February 6, 2014 at 3:18pm*

**Math - Trig**

cos C=0 and you just have the Pythagorean Theorem.
*Thursday, February 6, 2014 at 5:15am*

**Math - Trig**

since the sides are all equal, s^2 = s^2+s^2 - 2s^2 cosθ cosθ = 1/2 since you can pick any side, the value for θ is good for all three angles.
*Thursday, February 6, 2014 at 5:15am*

**Math - Trig**

Think about an obtuse triangle ABC where we know b and c and A. Then a/sinA = c/sinC allows for C to be acute or obtuse
*Thursday, February 6, 2014 at 5:11am*

**Math - Trig**

Explain why we cannot solve an oblique triangle with the Law of Sines given SAS.
*Thursday, February 6, 2014 at 3:37am*

**Math - Trig**

Use the law of cosines to show that the measure of each angle of an equilateral triangle is 60deg. Explain your reasoning.
*Thursday, February 6, 2014 at 2:54am*

**Math - Trig**

What familiar formula can you obtain when you use the third form of the Law of Cosines c^2 = a^2 + b^2 - 2ab cos C and you let C = 90deg? What is the relationship between the Law of Cosines and this formula?
*Thursday, February 6, 2014 at 2:54am*

**trig**

d/152 = cot 80°
*Thursday, February 6, 2014 at 12:04am*

**trig**

An advertising blimps hovers over stadium at the altitude of 152 m.the pilot sites a tennis court at in 80 degree angle of depression. Find the ground distance in the straight line between the stadium and the tennis court. (note: in an exercise like this one, and answers ...
*Thursday, February 6, 2014 at 12:04am*

**trig**

An advertising blimps hovers over stadium at the altitude of 152 m.the pilot sites a tennis court at in 80 degree angle of depression. Find the ground distance in the straight line between the stadium and the tennis court. (note: in an exercise like this one, and answers ...
*Thursday, February 6, 2014 at 12:00am*

**trig**

h/80 = sin40
*Wednesday, February 5, 2014 at 11:51pm*

**trig**

A kite is flying at an angle of elevation of about 40 degrees. All 80 meters of string have been let out. Ignoring the sag in the string, find the height of the kite nearest ten meters.
*Wednesday, February 5, 2014 at 11:51pm*

**trig**

A kite is flying at an angle of elevation of about 40 degrees. All 80 meters of string have been let out. Ignoring the sag in the string, find the height of the kite nearest ten meters.
*Wednesday, February 5, 2014 at 11:49pm*

**math**

have you no calculator? If not just enter it into google's search box Or, there are any number of online trig calculators
*Wednesday, February 5, 2014 at 6:00am*

**trig**

Just supposing you meant what you typed cot^2x+csc=x cos^2 x/sin^2 x + 1 sinx = x wolfram shows 4 answers in your domain http://www.wolframalpha.com/input/?i=solve+cot%5E2%28x%29+%2B+csc%28x%29+%3D+x x = appr. 1.211, 2.343, 3.523 and 5.964 When substituted all are ...
*Tuesday, February 4, 2014 at 8:22pm*

**trig - typos**

Well, you're not gong to make it easy, are you? csc(?) also, setting a trig function(x) = x is not amenable to solution by hand. So, assuming you meant cot^2(x)+csc(x) = 0, since cot^2 = csc^2-1, csc^2(x)-1+csc(x) = 0 csc(x) = (-1 +/- sqrt(5))/2 still not a usually used ...
*Tuesday, February 4, 2014 at 7:15pm*

**trig**

Find all the solutions from [0, 2pi] cot^2x+csc=x
*Tuesday, February 4, 2014 at 7:08pm*

**Trig**

There are 2pi radians per revolution so, 3rev/s * 2pi rad/rev = 6pi rad/s
*Monday, February 3, 2014 at 5:45am*

**Trig**

A fan makes 3 revolutions per second. The blades are 21 inches long. How do I find the angular velocity of a fan blade? in radians per second
*Monday, February 3, 2014 at 4:24am*

**trig - dang it!**

think twice, type once... 100/w = tan 79°
*Monday, February 3, 2014 at 12:23am*

**trig**

w/100 = tan 79°
*Monday, February 3, 2014 at 12:22am*

**trig**

Your search for A surveyor is trying to determine the width of a river. He forst stands at point A, directly opposite a tree at point B. He then walks 100 feet to point C. He measures the acute angle at point C to be 79 degrees. What is the width of the river?
*Sunday, February 2, 2014 at 8:56pm*

**Trig**

oh! thank you!
*Saturday, February 1, 2014 at 11:36pm*

**Trig**

sin Ø = .6 = 6/10 = 3/5 you should recognize the famous 3-4-5 right-angled triangle. so in quadrant II, construct your right-angled triangle in quadrant II sinØ = opposite/hypotenuse = y/r y = 3, r = 5 and in II, x = -4 so ... sinØ = 3/5 , csc Ø = 5...
*Saturday, February 1, 2014 at 11:12pm*

**Trig**

the sine of an angle is 0.6. the constraint is the angle lies in quadrant II. how do I figure out what x is to solve the six functions?
*Saturday, February 1, 2014 at 10:28pm*

**trig**

sin < 0 in QIII,QIV sec < 0 in QII,QIII so, what do you think?
*Thursday, January 30, 2014 at 6:18am*

**trig**

if sin and sec of theta are less than zero what quadrant is it in
*Wednesday, January 29, 2014 at 7:11pm*

**trig**

√3/2 ka man
*Wednesday, January 29, 2014 at 6:10am*

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