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April 19, 2014

April 19, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**Math-trig**

Wait. I see it. One face of the pyramid is an isosceles triangle with base angles of 70° So, the altitude of one of the faces is 92 sin70°, and half the base of one face is 92 cos70° so, if h is the pyramid height, sin52° = h/(92sin70°) h = 68.12 or, tan52&...
*Wednesday, December 19, 2012 at 4:48pm*

**Math-trig**

I can't see how this makes sense. Apparently the staircase runs up along one edge of the pyramid. Since the edge is longer than the altitude of one of the faces, its angle of elevation must be less. Better describe the diagram. Be sure to mark and label all points and ...
*Wednesday, December 19, 2012 at 4:40pm*

**Math-trig**

A stairway runs up the edge of the pyramid. From bottom to top the stairway is 92 m long. The stairway makes an angle of 70° to the base edge, as shown. A line from the middle of one of the base edges to the top of the pyramid makes an angle of elevation of 52° with ...
*Wednesday, December 19, 2012 at 4:33pm*

**trig**

1 - sec^2 = (cos^2 -1)/cos^2 = -tan^2 Divide that by tan^2 and you get -1
*Wednesday, December 19, 2012 at 1:31am*

**trig**

simplify (1-sec^2(è)/(tan^2(è)
*Wednesday, December 19, 2012 at 12:01am*

**Grade 11 math**

plot (6,8) join to the origin (0,0) from (6,8) draw a perpendicular to the x-axis to get a right-angled triangle , with the line from the origin to the point (6,8) as the hypotenuse. If you are studying trig you must have learned the Pythagorean Theorem let the length of the ...
*Tuesday, December 18, 2012 at 9:15pm*

**Grade 11 math**

based on a right-angled triangle , you probably originally learned the following definitions of trig functions: sinØ = opposite/hypotenuse cosØ = adjacent/hypotenuse tanØ = opposite/adjacent by now I hope you also learned them in the following terms sin&...
*Tuesday, December 18, 2012 at 8:51pm*

**trig/precalc**

actually, that's heading, not bearing. Adding up the x- and y-components of the two speeds, you get (507.1,531.2) so, the speed is 734.4 mph at N43.7°E
*Sunday, December 16, 2012 at 6:03pm*

**trigonometry**

recall that most fundamental of all trig identities sin^2 + cos^2 = 1 burn that into your brain in letters of fire. It is indispensable!
*Sunday, December 16, 2012 at 5:41pm*

**trig/precalc**

An airplane with a ground speed of 750 mph and a bearing of N 40 E encounters a wind of 50 mph with a bearing of S 30 E. Find the bearing and ground speed of the plane.
*Sunday, December 16, 2012 at 5:35pm*

**college trig word problem**

ty
*Saturday, December 15, 2012 at 2:37am*

**college trig word problem**

A Ferris wheel has a radius of 25 feet.The wheel is rotating at two revolutions per minute.Find the linear speed, in feet per minute, of a seat on this ferris wheel.
*Saturday, December 15, 2012 at 2:16am*

**trig**

if tan x= 4/3 and pip < x < 3pi/2, and cot y= -5/12 with 3pi/2 < y < 2pi find sin(x-y)
*Thursday, December 13, 2012 at 9:14pm*

**trig**

evaluate the following in exact form, where the angeles alpha and beta satisfy the conditions: sin alpha=4/5 for pi/2 < alpha < pi tan beta=7/24 for pie < beta < 3pi/2 answer choices A. sin(beta+alpha) B. tan(beta-alpha) C. cos(alpha-beta)
*Thursday, December 13, 2012 at 9:09pm*

**Trig**

15^(-x-3)=-17^(-3x) using a log base 10
*Wednesday, December 12, 2012 at 6:36pm*

**trig**

V = 425mi/h @ 63o + 31mi/h @ 24o X = 425*cos63 + 31*cos24 = 221.3 mi/h. Y = 425*sin63 + 31*sin24 = 391.3 mi/hr. tanA = Y/X = 391.3/221.3 = 1.7681 A = 60.5o. = Direction of plane. V = X/cosA = 221.3/cos60.5 = 431 mi/h = Velocity of plane.
*Tuesday, December 11, 2012 at 9:08pm*

**Adv. Geometry**

In general, if each side is x then area = 6(1/2)(x)(√3x/2) =(3/2)√3 x^2 find area of one of the 6 equilateral triangles the base will be x , and using Pythagoras or basic trig it can be show that the height is √3x/2
*Tuesday, December 11, 2012 at 8:43pm*

**precalc h**

sin(theta)/1-cos(theta) + 1-cos(theta)/sin(theta) = 2csc(theta) That question makes absolutely no sense.. could someone help me? or lead me in the direction to figuring it out? and.. (Beside the trig functions is theta) 1+1/cos = tan^2/sec-1
*Sunday, December 9, 2012 at 8:17pm*

**trig**

An airplane flies with a speed of 425 mph and a heading of 63°. If the heading of the wind is 24°and the speed of the wind is 31 mph, what is the heading of the plane and the ground speed? I've done a couple questions like this but still a little fuzzy on setting ...
*Sunday, December 9, 2012 at 11:01am*

**maths**

I assume you mean sin x/√(4+cos^2 x) let u = cos x du = -sin x dx and you have -∫1/√(4+u^2) du standard hyperbolic trig substitution is used to get -arcsinh(u/2) = -arcsinh(cosx / 2)
*Saturday, December 8, 2012 at 4:46pm*

**trig**

100cos10=x 100sin10=y <x,y>
*Friday, December 7, 2012 at 9:53am*

**trignometry**

make yourself a sketch of the 30-60-90 ° and the 45-45-90 ° right angled triangles the ratios of those sides are (in the same order) 1--√3--2 and 1--1--√2 so sin30 = 1/2 --- csc30 = 2 cos30 = √3/2 --- sec30 = 2/√3 tan30 = 1/√3 ----cot 30...
*Thursday, December 6, 2012 at 8:44am*

**Trig**

Cos (cos/sin)= cos^2/sin = (1 - sin^2x)/sin 1/sin sin^2/sin = csc x sin x
*Tuesday, December 4, 2012 at 11:35pm*

**Trig**

What are we doing? Solving ? or proving it is an identity? BTW, you have to put the x for all the trig functions that is, cosx cotx = cscx - sinx Let's see if it is an identity... LS = cosx (cosx/sinx = cos^2 x/sinx RS = 1/sinx - sinx = (1 - sin^2 x)/sin = cos^2 x/sinx = ...
*Tuesday, December 4, 2012 at 11:33pm*

**Trig**

cos x cot = csc - sin
*Tuesday, December 4, 2012 at 11:27pm*

**trigonometry**

can i use factoring to simplify this trig identity? the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer . this is the step i went through: 1) sinx...
*Monday, December 3, 2012 at 11:52pm*

**trigonometry**

1) Perform the operation and leave the result in trig. form. [3/4(cos pi/3 + i sin pi/3)][4(cos 3pi/4 + i sin 3pi/4)] Thanks
*Monday, December 3, 2012 at 10:28pm*

**Trig**

The height of a rider on a Ferris wheel is given by h(t)=12-10c0s(2pi)(t) meters, where t gives time, in minutes of the ride. a. Find the amplitude, midline, and period of the function h. b. During the first two minutes of the ride, find the times when the rider has a height ...
*Monday, December 3, 2012 at 7:09pm*

**Trig**

A population of animals oscillates annually from a low of 1300 on January 1st to a high of 2200 on July 1st, and back to a low of 1300 on the following January. Assume that the population is well-approximated by a sine or a cosine function. a. Find a formula for the population...
*Monday, December 3, 2012 at 7:06pm*

**trig**

Since we are doubling, let's use a base of 2 so 1000 (2)^t = 600,000 2^t = 600 log 2^t = log 600 tlog 2 = log 600 t = log 600/log 2 = appr 9.23 days number of 7 days 600,000 = n 2^6 n = 600000/2^7 = 4687.5 I would purchase 5000
*Monday, December 3, 2012 at 1:25pm*

**trig**

A 600,000-cell bacteria culture is needed for a lab experiment, but you only could purchase 1,000 cells. If the cells double each day, exactly how long will it be before the experiment can be run? Create a model for the bacteria. If the experiment is in seven days, how many ...
*Monday, December 3, 2012 at 12:51pm*

**Trigonometry**

sin 75 = .5 BC/15 BC = 30 sin 75 and sorry used trig (have no idea how to do it without trig) to get exact .5 BC/15 = cos 15 so BC = 30 cos 15 cos 15 = sqrt ( [1+cos 30]/2 ) but cos 30 = .5 sqrt 3 so BC = 30 sqrt ( [ 1 + .5 sqrt 3]/2 )
*Saturday, December 1, 2012 at 9:26pm*

**Calculus**

It took me your trig substitution followed by a u substitution to get 93squrt(x^2+9)/9x+C
*Saturday, December 1, 2012 at 1:47am*

**Calculus**

using trig substitutions, you can show that integral sqrt(a^2 + x^2) = 1/2 (x*sqrt(a^2+x^2) + a^2 log(x + sqrt(a^2+x^2))) = arcsinh(x/a)
*Saturday, December 1, 2012 at 12:11am*

**trig**

draw a diagram. You will see that (h + c sin b)/(c cos b) = tan a so, (h + 235 sin 18°)/(235 cos 18°) = tan 48° h = 175.6 ft
*Thursday, November 29, 2012 at 2:21pm*

**trig**

A tree on a hillside casts a shadow c ft down the hill. If the angle of inclination of the hillside is b to the horizontal and the angle of elevation of the sun is a, find the height of the tree. Assume a = 48°, b = 18° c = 235 ft. (Round your answer to the nearest ...
*Thursday, November 29, 2012 at 1:05pm*

**trig**

I assume you are studying vectors. Draw a line at 28° of length 200 (your resultant) Draw a line at 38° of length 207, join its tail to the first line. That smaller line will be the vector representing the wind I see a triangle with sides 200 and 207, with a contained ...
*Thursday, November 29, 2012 at 8:40am*

**trig**

measured from the plane, the building subtends an angle of 9° the distance d from the top of the building to the plane is found by d/sin50° = 40/sin9° d = 195.87 now, knowing d, the height h of the plane can be found: (h-40)/d = sin31° (h-40)/195.87 = sin31°...
*Thursday, November 29, 2012 at 5:28am*

**precalc/trig**

that would be cosx cos pi/4 - sinx sin pi/4 + cosx cos pi/4 + sinx sin pi/4 = 1 2cosx cos pi/4 = 1 √2 cosx = 1 cosx = 1/√2 x = pi/4, 7pi/4
*Thursday, November 29, 2012 at 5:18am*

**precalc/trig**

olve for the interval [0, 2pi]. cos(x+ pi/4)+cos(x- pi/4)=1
*Thursday, November 29, 2012 at 1:25am*

**trig**

A plane has an airspeed of 200 miles per hour and a heading of 28.0°. The ground speed of the plane is 207 miles per hour, and its true course is in the direction of 38.0°. Find the speed and direction of the air currents, assuming they are constants. (Round your ...
*Thursday, November 29, 2012 at 12:24am*

**trig**

angles of elevation to an airplane are measured from the top and the base of a building that is 40 m tall. The angle from the top of the building is 31°, and the angle from the base of the building is 40°. Find the altitude of the airplane. (Round your answer to two ...
*Wednesday, November 28, 2012 at 11:43pm*

**trig**

Should this be Q IV (quadrant 4) tan-1(-4/3) = -53.13 degrees (-53.13 degrees = 360 + -53.13 = 306.87, is in quadrant 4) Evaluate with your calculator cos(2*306.87)
*Wednesday, November 28, 2012 at 3:19pm*

**calculus**

Use some trig identities: tan(-x) = tan(x) csc(x) = 1/sin(x) tan(x) = sin(x)/cos(x) tan(-x)*csc(x) = tan(x)*(1/sin(x) = (sin(x)/cos(x))*(1/sin(x) = 1/cos(x) = sec(x)
*Wednesday, November 28, 2012 at 3:15pm*

**trig**

cosθ = 2/√7 and 0 <= θ <= π means we are in the 1st quadrant. so, sinθ = √(7-4)/√7 = √(3/7)
*Wednesday, November 28, 2012 at 2:59pm*

**trig**

given that 0 is less then or equal to theta and theta is less then or equal to pi. ans cos theta = 2/squar root of 7. Find sin theta
*Wednesday, November 28, 2012 at 2:21pm*

**Pre-Cal Trig**

Two lookout towers are situated on mountain tops A and B, 4 mi from each other. A helicopter firefighting team is located in a valley at point C, 3 mi from A and 2 mi from B. Using the line between A and B as a reference, a lookout spots a fire at an angle of α = 37° ...
*Wednesday, November 28, 2012 at 12:20pm*

**Trigonometry **

How do you solve this trig identity? Using @ as theta: Sin@-1/sin@+1 = -cos@/(sin@+1)^2 I've tried it multiple times but I can't seem to arrive at the answer.
*Tuesday, November 27, 2012 at 11:41pm*

**trig**

well, you know that siny = 3/5 cos2y = 1 - 2sin^2 y or, you know that tan 2y = 2tany/(1-tan^2 y) and sec^2 2y = 1 + tan^2 2y good way to check your answer.
*Tuesday, November 27, 2012 at 11:29pm*

**trig**

If tan y = -4/3 and y is in Q I'VE find cos 2y
*Tuesday, November 27, 2012 at 11:26pm*

**Math (trigonometry)**

How do you solve this trig identity problem without factoring? I just used @ to represent theta: sin^4@ + 2sin^2@cos^2@ + cos^4@ = 1
*Tuesday, November 27, 2012 at 11:25pm*

**trig**

If tan y = -4/3 and y is in Q I'VE find cos 2y
*Tuesday, November 27, 2012 at 11:21pm*

**calculus**

cot(pi/2-x) = tan(x) that what you mean? In fact, for any trig function f, co-f(x) = f(pi/2-x). The "co" means complementary angle.
*Tuesday, November 27, 2012 at 10:20am*

**Trigometry**

This is not trig ... it is basic algebra but anyway (2x-1)(x+2)(3x-4)-(4x^2m+7x-2) is simplified to 6x^3+17x^2-25x+10 is a false statement, since the first part contains a variable m, which has disappeared in the last part of the statement. Unless your statement contains a ...
*Monday, November 26, 2012 at 10:57pm*

**Trigometry**

yes , this is trig. Can you please help me
*Monday, November 26, 2012 at 9:33pm*

**Trigometry**

yes , this is trig. Can you please help me
*Monday, November 26, 2012 at 9:31pm*

**Trigometry**

Is there a question here? Is this Trig?
*Monday, November 26, 2012 at 9:18pm*

**trig**

i think i figured out my mistake. it should be 720 = 713.7+152.7 sin [(2pi/366)(t-80.75)] this problem is modelling minutes of daylight in a town using the form y = d+a sin[b(t-c)]. let y equal 720 minutes of daylight and let t=the number of days in 2012
*Monday, November 26, 2012 at 2:51am*

**trig**

It is intuitively obvious that this equation cannot have a solution the way it was written the sin(anything) has to be a number between -1 and +1 so if 720 = 11.895 + x , the value of x cannot possible fall between - 1 and +1 check your typing, or if correctly typed, there is ...
*Sunday, November 25, 2012 at 8:35pm*

**trig**

same way you would solve for any equation. collect t stuff on one side, and work your way inside out the expressions till t ia all alone: 708.105 = 2.545sin [2pi/366(t-80.5)] 278.2338 = sin [2pi/366(t-80.5)] at this point you are stuck. sin(x) is never more than 1, so there is...
*Sunday, November 25, 2012 at 8:32pm*

**trig**

How would I solve this problem for t? 720 = 11.895+2.545sin [2pi/366(t-80.5)]
*Sunday, November 25, 2012 at 5:36pm*

**trig**

what you want to do is to show that using standard identities and algebraic manipulation, the left side can be made equal to the right side. cosx(tanx+cotx) change to sin,cos using trig identities cosx(sinx/cosx + cosx/sinx) now do some algebra cosx*sinx/cosx + cosx*cosx/sinx ...
*Sunday, November 25, 2012 at 2:58pm*

**trig**

I don't understand what to do after i convert everything into cos and sin. Verify: cosx(tanx+cotx)=cscx
*Sunday, November 25, 2012 at 2:34pm*

**Math**

The trig functions are ratios of x, y and the hypotenuse sqrt (x^2+y^2). Therefore if you take the angle between the ray and the x axis you can figure out the sin, cos and tan. The ratios are the same in every quadrant but the signs are not. They are also the same using (90-...
*Sunday, November 25, 2012 at 12:32pm*

**Math**

Am I missing something? I think it's a little different because there's a 150 degree angle that starts in quadrant one and goes to quadrant two, called the principal angle, and the left over 30 degree angle made by the original 150 degree angle is the related acute ...
*Sunday, November 25, 2012 at 9:17am*

**Math**

Sure, just like we were doing with your 110 degree problem. That is 20 degrees from the y axis or 70 degrees from the x axis in quadrant 2 draw the other three lines that are 20 degrees from y and 70 degrees from x axes in the other three quadrants. 70 , 110 , 350 , 290 These ...
*Sunday, November 25, 2012 at 6:45am*

**Math**

the thing is that 70 degrees, 110 degrees, 250 degrees and 290 degrees are all 20 degrees from the vertical axis and have the same absolute values of trig functions HOWEVER the signs depend on the quadrant
*Saturday, November 24, 2012 at 11:01pm*

**Math**

Sorry, I still don't understand... I know I seem really dumb right now, but this unit just isn't working for me for some reason.. I'm usually pretty decent at math. We just started grade 11 trig and I'm totally lost after the first lesson. I understand that it...
*Saturday, November 24, 2012 at 10:53pm*

**Pre-Calculus**

one period is from x = 0 to x = 2 pi/3 so that the argument of the trig function changes by 2 pi so the answer is 2 pi/3
*Saturday, November 24, 2012 at 6:38pm*

**math**

Simplify sin 2 theta over 2 cos theta to a single primary trig function
*Thursday, November 22, 2012 at 9:51pm*

**trig**

cos theta = sqrt ( 3 ) / 2 for theta = pi /6 and and theta = 11 pi / 6 Solutions : 3 x - pi / 2 = pi / 6 3 x = pi / 6 + pi / 2 3 x = pi / 6 + 3 pi / 6 3 x = 4 pi / 6 3 x = 2 pi / 3 Divide both sides by 3 x = 2 pi / 9 AND 3 x - pi / 2 = 11 pi / 6 3 x = 11 pi / 6 + pi / 2 3 x = ...
*Wednesday, November 21, 2012 at 5:05am*

**math**

cos(3x- ƒÎ/2) = sin(3x) (Check your trig identities) 3x = sin^-1(-ã3/2) = 4 pi/3, 5 pi/3, 10 pi/3, 11 pi/3 etc x = 4 pi/9 etc
*Wednesday, November 21, 2012 at 1:01am*

**Physics**

he graph of the velocity of a mass attached to a horizontal spring on a horizontal frictionless surface as a function of time is shown below. The numerical value of V is 5.48 m/s, and the numerical value of t0 is 8.97 s. a) What is the amplitude of the motion in m? b) What is ...
*Tuesday, November 20, 2012 at 11:47pm*

**trig**

find all the values of X between 0 and 2π such that cos(3x-π/2) = √3/2
*Tuesday, November 20, 2012 at 11:18pm*

**TRIG HELP**

you're kidding, right? tan pi/4 = 1 x = k*pi + pi/4 for any integer k
*Tuesday, November 20, 2012 at 10:41am*

**TRIG HELP**

find the exact solutions over the interval tanx = 1 all real x
*Tuesday, November 20, 2012 at 10:35am*

**math**

(-3/5, 2) is a point on the terminal side of theta, find the value of the six trig functions
*Monday, November 19, 2012 at 9:56pm*

**trig**

If the ship has sailed to point C, angle A = 73°-31° = 42° angle B = 17° so, angle C is 121° the distance AC is side a, where a/sin42° = 4.7/sin121° a = 3.67
*Monday, November 19, 2012 at 10:19am*

**trig**

It is 4.7km from Lighthouse A to Port B. The bearing of the port from the lighthouse is N73E. A ship has sailed due west from the port and its bearing from the lighthouse is N31E. How far has the ship sailed from the port?
*Sunday, November 18, 2012 at 11:38pm*

**trig**

I am not familiar with your symbols. sinB/b = sinA/a. sinB/23.1 = sin115.2/43.6 Multiply both sides by 23.1 sinB = 23.1*sin115.2/43.6 = 0.47939 B = 28.6o. CosA = b^2+c^2-a^2/2bc. Cos115.2 = (23.1)^2+c^2-(43.6)^2/46.2c -0.42578 = (533.61+c^2-1901)/46.2 -0.42578 = (-1367.39+c^2...
*Sunday, November 18, 2012 at 9:35pm*

**trig**

V = 16Knots @ 135o+15Knots @ 195o. X = 16*cos135 + 15*cos195 = -25.80 Knots Y = 16*sin135 + 15*sin195 = 7.43 Knots. tanA = Y/X = 7.43/-25.80 = -0.28798 Ar = -16.07o = Reference angle. A = -16.07 + 180 = 164o = Direction. V = X/cosA = -25.80/cos164 = 27 Knots.
*Sunday, November 18, 2012 at 8:04pm*

**trig--power reducing formulas**

(cosx)+(cosx)(cosx)
*Sunday, November 18, 2012 at 7:01pm*

**trig--power reducing formulas**

(cosx)+(cosx)(cosx)
*Sunday, November 18, 2012 at 7:00pm*

**Trig**

if the boat travels s seconds, 1 km/hr = 1.4667 ft/s 209^2 + (6*1.4667s)^2 = (32*1.4667s)^2 s = 4.5 sec
*Sunday, November 18, 2012 at 3:24pm*

**Trig**

If it takes x hours to cross the river as intended, and 200ft = .06096 km 0.06096^2 + (5x)^2 = (35x)^2 x = 0.00175976 tanθ = 5*0.00175976 / 0.06096 = 0.14433 θ = 8.2° hmm. or, with less calculation, sinθ = 5/35 = 1/7 θ = 8.2°
*Sunday, November 18, 2012 at 3:13pm*

**Trig**

Suppose you would like to cross a 200-foot wide river in a boat. Assume that the boat can travel 35 mph relative to the water and that the current is flowing west at the rate of 5 mph. What bearing should be chosen so that the boat will land at a point exactly across from its ...
*Sunday, November 18, 2012 at 5:17am*

**Trig**

Suppose you would like to cross a 209-foot wide river in a boat. Assume that the boat can travel 32 mph relative to the water and that the current is flowing west at the rate of 6 mph. If the bearing chosen is chosen so that the boat will land at a point exactly across from ...
*Sunday, November 18, 2012 at 5:12am*

**Physics**

The argument of the trig function can be written several ways, for example (2 pi t/T), the way I did it because you were given T or (2 pi f t) because T = 1/f or ( w t) where w is the Greek letter omega and is the radial frerquency, 2 pi f, in radians/second in any case, when ...
*Saturday, November 17, 2012 at 5:54pm*

**trig**

there is redundant (and slightly inconsistent) information the distance is 52 cos57° = 28.32 √(52^2 - 44^2) = 27.71
*Friday, November 16, 2012 at 10:35am*

**trig - incomplete**

where's D? is ABC a triangle?
*Friday, November 16, 2012 at 10:32am*

**trig**

150 at 155° = 63.4i -135.9j 25 at 160° = 8.6i - 23.5j sum = 72.0i - 159.4j = 175.0 at 155.7°
*Friday, November 16, 2012 at 10:30am*

**trig**

A ship is headed due north at a constant 20 miles per hour. Because of the ocean current, the true course of the ship is 15°. If the currents are a constant 18 miles per hour, in what direction are the currents running? (Enter your answers as a comma-separated list. Round ...
*Friday, November 16, 2012 at 6:30am*

**trig**

A 52-foot wire running from the top of a tent pole to the ground makes an angle of 57° with the ground. If the length of the tent pole is 44 feet, how far is it from the bottom of the tent pole to the point where the wire is fastened to the ground? (The tent pole is not ...
*Friday, November 16, 2012 at 6:30am*

**trig**

Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest whole number. Round your answers for sides c and c' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank...
*Friday, November 16, 2012 at 6:30am*

**trig**

Find all solutions to the following triangle. (Round your answers for angles A, C, A', and C' to the nearest minute. Round your answers for sides a and a' to two decimal places. If either triangle is not possible, enter NONE in each corresponding answer blank.) B...
*Friday, November 16, 2012 at 6:29am*

**trig**

Find all solutions to the following triangle. (Round your answers for angles A, B, A', and B' to the nearest minute. Round your answers for sides a and a' to the nearest whole number. If either triangle is not possible, enter NONE in each corresponding answer blank...
*Friday, November 16, 2012 at 6:29am*

**trig**

Find all solutions to the following triangle. (Round your answers to one decimal place. If either triangle is not possible, enter NONE in each corresponding answer blank.) A = 115.2¡ã, a = 43.6 cm, b = 23.1 cm First triangle (assume B ¡Ü 90¡&...
*Friday, November 16, 2012 at 6:28am*

**trig**

Given BC = 53 cm, BD = 62 cm, CD = 80 cm, ABC = 53°, and ACB = 66°, find the following. (Round your answers to the nearest whole number.) (a) the length of the chainstay, AC AC = cm (b) BCD BCD = °
*Friday, November 16, 2012 at 6:20am*

**trig**

A barge is pulled by two tugboats. The first tugboat is traveling at a speed of 16 knots with heading 135°, and the second tugboat is traveling at a speed of 15 knots with heading 195°. Find the resulting speed and direction of the barge. (Round your answers to the ...
*Friday, November 16, 2012 at 6:19am*

**trig**

A plane has an airspeed of 190 miles per hour and a heading of 24.0°. The ground speed of the plane is 214 miles per hour, and its true course is in the direction of 40.0°. Find the speed and direction of the air currents, assuming they are constants. (Round your ...
*Friday, November 16, 2012 at 6:18am*

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