Monday

April 21, 2014

April 21, 2014

**Recent Homework Questions About Trigonometry**

Post a New Question | Current Questions

**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*

**trig**

A plane is flying with an airspeed of 150 miles per hour and heading 155°. The wind currents are running at 25 miles per hour at 160° clockwise from due north. Use vectors to find the true course and ground speed of the plane.
*Friday, November 16, 2012 at 6:17am*

**trig**

200 200.5 201.0 ... h(x) = 200 + 0.5x
*Friday, November 16, 2012 at 12:13am*

**trig**

A certain species of tree grows an average of 0.5 cm per week. Write an equation for the sequence that represents the weekly height of this tree in centimeters if the measurements begin when the tree is 200 centimeters tall.
*Thursday, November 15, 2012 at 10:03pm*

**trig**

since we're in QII, sinθ = 1/5 cosθ = -√24/5 now, just plug into the formulas. what do you get?
*Thursday, November 15, 2012 at 3:51pm*

**trig**

given that 1/2pi<theta<pi and sin theta=1/5, use appropriate trigonometric formulas to find the exact values of the following (i) cos(2theta) (ii) cos theta (iii) sin(2theta)
*Thursday, November 15, 2012 at 2:10pm*

**Physics**

Calculate the ratio of the kinetic energy to the potential energy of a simple harmonic oscillator when its displacement is 1/8 of its amplitude. (The answer is an integer.) Approach: Choose a specific trigonometric form for the position function x(t). It doesn't matter ...
*Thursday, November 15, 2012 at 2:00am*

**Difficult Trig Word Problem**

y = 4^(-3t) * sin(2πt) so if we set y = 0 4^(-3t) = 0 ----> so solution or sin(2πt) = 0 2πt = 0 or 2πt = π or 2πt = 2π or .. t = 0 or t = 1/2 or t = 1 or t= .. looks like every half second, starting with t = 0
*Wednesday, November 14, 2012 at 9:56pm*

**Difficult Trig Word Problem**

sorry y(t) =4e(then the exponent -3t) all times sin(2pi*t)
*Wednesday, November 14, 2012 at 6:43pm*

**Difficult Trig Word Problem**

I am not sure what the exponent is the way you typed it using no brackets, it would be (4e^-3) * tsin(2πt) or is it 4e^( -3tsin(2πt) )
*Wednesday, November 14, 2012 at 6:41pm*

**Difficult Trig Word Problem**

The displacement of a spring vibrating in damped harmonic motion is given by y(t) = 4e^-3t sin(2pi*t) where y = displacement and t = time with t greater than/equal to zero. Find the time(s) when the spring is at its equilibrium position (y=0). The number "e" is Euler...
*Wednesday, November 14, 2012 at 5:51pm*

**trig**

a ladder leaning against a vertical wall make an angle 24 degrees with the wall.the foot of the ladder is 5 ft from the wall find the length of the ladder
*Wednesday, November 14, 2012 at 3:56pm*

**trig**

sin 25° = 4.5/x x = 4.5/sin25 = appr 10.6 ft
*Tuesday, November 13, 2012 at 9:05pm*

**trig**

Olivia has a pool slide that makes an angle of 25º with the water. The top of the slide stands 4.5 feet above the surface of the water. The slide makes a straight line into the water. How long is the slide?
*Tuesday, November 13, 2012 at 8:08pm*

**Precalc**

LS = cos(-x) - sin(-x) = cosx + sinx = RS These are based on standard trig relationships that you should know and recognize
*Tuesday, November 13, 2012 at 10:18am*

**Trig Identity**

LS = sin^4 Ø - cos^4 Ø = (sin^2 Ø + cos^2 Ø)(sin^2 Ø - cos^2 Ø) = (1)(sin^2 Ø - cos^2 Ø) = RS
*Tuesday, November 13, 2012 at 10:10am*

**Trig Identity**

Prove or disprove the following identity sin^4theta - cos^4theta = sin^2theta-cose^2theta
*Tuesday, November 13, 2012 at 9:46am*

**trig**

or taking it at face value the way it was typed tan 2A - 2tan A = -1 2tanA/(1-tan^2 A) - 2tanA = -1 let tanA = x for easier typing 2x/(1-x^2) - 2x = -1 times 1 - x^2 2x - 2x + 2x^3 = -1 + x^2 2x^3 - x^2 + 1 = 0 by Wolfram http://www.wolframalpha.com/input/?i=2x%5E3+-+x%...
*Monday, November 12, 2012 at 5:52pm*

**trig**

Hmmm. Sensing a missing ^ character, we have tan^2(A) - 2tanA + 1 = 0 (tanA - 1)^2 = 0 tanA = 1 So, A = 45º or 225º
*Monday, November 12, 2012 at 5:26pm*

**trig**

Solve tan2A - 2tanA = -1 for 0º≤A≤360º.
*Monday, November 12, 2012 at 5:16pm*

**trig**

Mmmhh, 5 consecutive trig questions without any work on your part, or any indication where you are having problems. Looks like homework dumping to me. Hint: A very effective way to prove identities is to change everything you see into sines and cosines using basic identities. ...
*Monday, November 12, 2012 at 7:53am*

**TRIG**

in this case you must use brackets or else the statement is false LS = cosx/(1+sinx) + (1+sinx)/cosx = cosx/(1+sinx) + (1+sinx)/cosx = ( cosx(cosx) + (1+sinx)(1+sinx) )/(cosx(1+sinx)) = (cos^2 x + 1 + 2sinx + sin^2 x)/(cosx(1+sinx)) = (1 + 1 + 2sinx)/(cosx(1+sinx)) = 2(1+sinx...
*Monday, November 12, 2012 at 7:50am*

**TRIG**

cotx = 5cosx cosx/sinx - 5cosx = 0 cosx(1/sinx - 5) = 0 cosx = 0 or 1/sinx = 5 if cosx = 0, x = π/2 , 3π/2 if 1/sinx = 5 sinx = 1/5 x = .20136 or 2.94023
*Monday, November 12, 2012 at 7:42am*

**trig**

cot^2/1 + cot^2 = cos^2
*Monday, November 12, 2012 at 7:35am*

**TRIG**

Sinxtanx = sec x
*Monday, November 12, 2012 at 7:34am*

**TRIG**

(cotx + cscx(1-cosx)= SIN x
*Monday, November 12, 2012 at 7:33am*

**TRIG**

Verify cos x/1+sinx + 1+ sinx/cosx = 2sec
*Monday, November 12, 2012 at 7:32am*

**TRIG**

Find one numerical value for cot x/cos x = 5
*Monday, November 12, 2012 at 7:31am*

**trig**

I assume you meant tan^2x/(1 + tan^2x) LS = tan^2x/(1 + sec^2x - 1) = tan^2x/sec^2 = (sin^2x/cos^2x) (cos^2x) = sin^2 x = RS
*Sunday, November 11, 2012 at 11:14pm*

**trig**

LS = secx - tanx + sinxsecx - sinxtanx = 1/cosx - sinx/cosx + sinx/cosx - sin^2x/cosx = 1/cosx - (1 - cos^x)/cosx = (1 - 1 + cos^2x)/cosx = cos^2x/cosx = cosx = RS
*Sunday, November 11, 2012 at 11:11pm*

**trig**

(1 + sinx)(secx - tanx) = cos x
*Sunday, November 11, 2012 at 9:52pm*

**trig**

tan^2x/1 + tan^2x = sin ^2 x
*Sunday, November 11, 2012 at 9:52pm*

**trig**

cos^2 A/ 1+sin A
*Sunday, November 11, 2012 at 9:51pm*

**precalc/Trig**

using the same addition formulas as before, note that 82.5 = 60 + 22.5 12 = 30-18
*Saturday, November 10, 2012 at 2:13am*

**precalc/Trig**

Assume we know the following: cos(22.5 degrees)= sqrt(2+sqrt2)/2 sin(22.5 degrees)=sqrt(2-sqrt2)/2 cos(18 degrees)=sqrt((sqrt5+5)/8) sin(18 degrees)=(sqrt5-1)/4 What is: Sin(82.5 degrees) cos(82.5 degrees) sin(12 degrees) cos(12 degrees)
*Saturday, November 10, 2012 at 2:11am*

**Precalc/Trig**

just plug in you addition formulas: sin(u+v) = sinu cosv + sinv cosu = (2/5)(1/3) + (√21/5)(√8/3) = 2(1+√42)/15 do the others likewise. If necessary, draw triangles and label the sides to evaluate the other functions of u,v,x,y
*Saturday, November 10, 2012 at 2:08am*

**Precalc/Trig**

evaluate the expression assuming that cos(x)=1/7, sin(y)=1/3, sin(u)=2/5 and cos(v)=1/3. what is cos(u+v)? sin(x-y)? and tan(u-v)?
*Saturday, November 10, 2012 at 12:48am*

**trig**

circumference of the potential whole circle = 2π(1.3) = 2.6π so , by a simple ratio arc/2.6π = 15/360 arc = 2.6π(15/360) = (13/120)π or appr .34 m or 34 cm
*Thursday, November 8, 2012 at 11:04pm*

**trig**

the pendulum of a grandfather clock is 1.3m long. Determine the length of the arc through which the pendulum moves if it moves through an angle of 15
*Thursday, November 8, 2012 at 9:11pm*

**Maths**

use the trig sum-to-difference formulas sinAcosB = (sin(A+B) + sin(A-B))/2 You must have seen such. If not a visit to http://oakroadsystems.com/twt/sumdiff.htm bears study
*Thursday, November 8, 2012 at 2:49pm*

**trig**

whoops, sorry
*Wednesday, November 7, 2012 at 2:52pm*

**Trig**

in QII, if csc a = 2, sin a = 1/2 cos a = -√3/2 tan a = -1/√3 if sec B = -3, cos B = -1/3 sin B = √8/3 tan B = -√8 Now it's just plug and chug, using your sum-of-angles and half-angle formulas. Come on back if you get stuck.
*Wednesday, November 7, 2012 at 11:32am*

**Trig **

Give: a=alpha B=beta csc a= 2 pi/2<a<pi secB=-3 pi/2<B<pi 1)find sin(a+B) 2)find tan(a-B) 3)cos B/2 4)sin a/2
*Wednesday, November 7, 2012 at 11:27am*

**trig - note**

sin is positive in QII
*Wednesday, November 7, 2012 at 11:22am*

**trig**

You are talking about 112.5 degrees that is 90 + 22.5 22.5 is half of 45 so draw this in the third quadrant I know functions of 45 degrees sin 45 = +1/sqrt 2 cos 45 = +1/sqrt 2 so what are the functions of 22.5 degrees? sin(45/2) = +/- sqrt[(1-cos 45)/2] cos (45/2) = +/-sqrt[(...
*Wednesday, November 7, 2012 at 11:17am*

**trig**

cos2θ = 1 - 2sin^2 θ since 2θ = 5pi/4, cos 2θ = -1/√2 -1/√2 = 1 - 2sin^2 5π/8 2sin^2 5π/8 = 1 + 1/√2 sin 5π/8 = √(1 + 1/√2)/2 = √(2+√2) / 2
*Wednesday, November 7, 2012 at 11:08am*

**trig**

How do you find the exact value of sin 5pi/8?
*Wednesday, November 7, 2012 at 10:51am*

**trig**

see whether you can figure how I developed the required formula for the height h of the tree: h = 121(cos8°tan50° - sin8°) = 126 ft Draw a diagram and figure out some of the distances.
*Tuesday, November 6, 2012 at 10:28am*

**TRIG**

How about the most obvious case, x - 0 LS = cos (π(0)) = cos 0 = 1 RS = 0(cos0) = 0(1) = 0 ≠ LS Now, a more interesting question would have been , For what values of x is the statement true? http://www.wolframalpha.com/input/?i=cos%28πx%29+%3D+π*cos...
*Tuesday, November 6, 2012 at 9:51am*

**TRIG**

Since the tangent is positive and the secant (or cosine) is negative, x must lie in quadrant III construct a triangle showing an opposite side of 1 and an adjacent side of √15 by Pythagoras, the hypotenuse is 4. so in III sinx = -1/4
*Tuesday, November 6, 2012 at 9:46am*

**TRIG**

show that cos(pi x) = pi cosx is not an identity by finding a single value of x for which it fails to hold.
*Tuesday, November 6, 2012 at 9:32am*

**TRIG**

Given that tanx = 1/sqr(15) and secx -4/sqr(15) find the value of sinx
*Tuesday, November 6, 2012 at 9:30am*

**trig**

A tree growing on a hillside casts a 121-foot shadow straight down the hill. Find the height of the tree (in feet) if the slope of the hill is 8 degrees and the angle of elevation of the sun from the horizontal is 50degrees.
*Monday, November 5, 2012 at 10:38pm*

**trig**

use the formula cos 2A = 1 - 2sin^2 A cos u = 1 - 2sin^2 (u/2) , π/2 < u < π , so u is in quad II -4/5 = 1 - 2sin^2 (u/2) 2 sin^2 (u/2) = 9/5 sin^2 (u/2) = 9/10 sin u/2 = √(9/10) = 3/√10 or 3√10/10
*Monday, November 5, 2012 at 12:27am*

**trig**

find the exact value of sinu/2, given that cosu=-4/5 and pi/2less than u less than pi
*Sunday, November 4, 2012 at 9:53pm*

**Pre Calculus**

you're welcome. I always enjoyed doing trig identities. They provide endless variations. The hard ones for me, though, are the ones involving tan and sec combinations. They can get quite involved.
*Sunday, November 4, 2012 at 1:54pm*

**physics**

distanchoriz=vh*timeinair but time in air can be found by hf=hi+vy*time-4.9t^2 or 0=0+vy*t-4.9t^2 0=t(vy-4.9) or time in air=vy/4.9 but vy=vsin25 and vh=vcos25 time in air=V*sin25/4.9 9.12=Vcos25*Vsin25/4.9-4.9(V *sin 25/4.9)^2 and you solve with V with some algebra and trig.
*Friday, November 2, 2012 at 10:01am*

**trig**

cos(x-13) = sqrt(3)/2 x-13 = pi/6 or 5pi/6 now you should be able to figure out x.
*Friday, November 2, 2012 at 1:36am*

**trig**

2 cos(theta-13)= square root of 3
*Friday, November 2, 2012 at 1:29am*

**Trig**

8.1
*Wednesday, October 31, 2012 at 11:34pm*

**TRIG HONORS CH: 5.8**

Hmm. You, NATO and Selena ought to be able to figure this out if you put your heads together...
*Tuesday, October 30, 2012 at 10:46am*

**trig**

just convert each distance/direction to x- and y-components, and add them up. Any problems?
*Tuesday, October 30, 2012 at 10:45am*

**TRIG**

I will assume that HR(1,-1) is a typo and you meant H(1,-1). Doesn't matter anyway, since it does not enter the calculation. The question reduces to a simple problem Rotae OG counterclockwise to OG' through and angle of 75° Clearly OG' has to be 10 , since OG...
*Tuesday, October 30, 2012 at 8:47am*

**TRIG**

Triangle FGH has vertices F(0, 10), G(10, 0), and HR(1, -1). After rotating triangle FGH counterclockwise about the origin 75º, the coordinates of G' to the nearest hundredth are (2.59, ?).
*Tuesday, October 30, 2012 at 1:53am*

**TRIG HONORS CH: 5.8**

a ship leaves port with a bearing of S 42deg W after traveling 7 miles, the ship turns 90deg and travels on a bearing of N 52deg W for 10 miles at that time, what is the bearing of the ship to the port?
*Monday, October 29, 2012 at 10:55pm*

**trig**

a ship leaves port with a bearing of S 42deg W after traveling 7 miles, the ship turns 90deg and travels on a bearing of N 52deg W for 10 miles at that time, what is the bearing of the ship to the port?
*Monday, October 29, 2012 at 10:53pm*

Pages: <<Prev | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | **13** | 14 | 15 | 16 | 17 | 18 | 19 | Next>>

Post a New Question | Current Questions