Number of results: 15,052
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 6:14pm by Steve
Trig
Should the triangle be solved beginning with Law of Sines of Law of Cosines. Then solve the triangle. Round to the nearest tenth. a=16, b=13, c=10. Cosines A=93 degrees, B=54 degrees, C=33 degrees
Friday, March 7, 2008 at 1:34pm by Jon
Trig. Law of Cosines
Show that any triangle with standard labeling... a^2+b^2+c^2/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c I don't get it. Can someone please help me. Start here with the law of cosines: a^2 = b^2 + c^2 -2bc Cos A b^2 = a^2 + c^2 -2ac Cos B c^2 = a^2 + b^2 - 2ab Cos C ...
Tuesday, December 12, 2006 at 5:45pm by ashley
Alge2/trig
Whether the law of cosines or the law of sones be used first to solve for the triangle
Saturday, March 14, 2009 at 10:54am by Sha
Trig word problem
since you know the height of the shadow and an angle, you can use a trig function to find the height of the flagpole, assuming that the triangle formed is a right triangle, unless you know the Law of Sines and the Law of Cosines.
Sunday, February 24, 2008 at 9:42pm by Laxmi
Trig word problem
since you know the height of the shadow and an angle, you can use a trig function to find the height of the flagpole, assuming that the triangle formed is a right triangle, unless you know the Law of Sines and the Law of Cosines.
Sunday, February 24, 2008 at 9:42pm by Laxmi
trig
I need to prove parallelogram law using the law of cosines. 2AB^2 + 2BC^2 = AC^2 + BD^2 Please, help me to do this.
Wednesday, December 7, 2011 at 12:18pm by Evelina
law of cosines
law of cosines m<c=60, a=12,b=15 triangle
Thursday, April 28, 2011 at 9:42pm by Anonymous
Trig
Use law of cosines: c^2=a^2 +b^2 -2abcosC C=38deg a=3.6*2 b=4.1*2
Monday, May 10, 2010 at 12:42pm by bobpursley
law of cosines
The law of cosines is: a^2 = b^2 + c^2 - 2cosA or b^2 = a^2 + c^2 - 2cosB or c^2 = a^2 + b^2 - 2cosC the lower case letters are sides, the capital letters are for angles. For this problem, you would do c^2 = 12^2 + 15^2 - 2cos60 That gives you side c. After that, you should be...
Thursday, April 28, 2011 at 9:42pm by Kris
Trig Help Please!!!
Find the indicated angle è. (Use either the Law of Sines or the Law of Cosines, as appropriate. Assume a = 95 and c = 137(angle B=38). Round your answer to two decimal places.)
Thursday, April 5, 2012 at 11:43pm by Adam
trig
Is this a triangle? Law of Cosines: a^2=b^2+c^2-2bcCosA solve for a. Then, law of sines: sin46/a=sinB/b solve for sinB, then B
Tuesday, June 29, 2010 at 3:59pm by bobpursley
trig
Presumably, a is opposite angle A, etc. By law of cosines, a^2 = b^2 + c^2 - 2bc cos A = 16 + 64 - 2*4*8*0.69 = 25.84 a = 5.98. Call it 6. So, by law of sines a/sin A = b/sin B sinB = bsinA/a = 4*.719/6 = .479 B = 28.6 deg
Friday, October 28, 2011 at 12:44am by Steve
Alge2/trig
If you are given three sides, start with the law of cosines to get one angle. Then you may go to the law of sines to get the second angle, then the third angle is solved by knowing the sum of the interior angles of a triangle is 180deg.
Saturday, March 14, 2009 at 10:54am by bobpursley
Trig
distance is the difference in positions. But that involves vectors, so here, I think your teacher wants you to use trig. draw the figure. Notice the angles first: the interior angle between the paths is 130degrees Next, the law of cosines c^2=20^2+15^2-2*20*15Cos130 that is it...
Wednesday, July 21, 2010 at 7:07pm by bobpursley
trig
This is just a straightforward law of cosines problem. We want to find a, given A,b,c. A=58°, c=7.5, b=8.6 a^2 = b^2+c^2 - 2bc cosA plug and chug. . .
Thursday, February 14, 2013 at 2:16pm by Steve
Math
You are dealing with the law of sines and cosines. I am not certain what you don't get. Is it the algebra? 1) b=14.2*sin67 / sin38 put these in the calculator and compute. 2) Same issue: a= 46*sin20/sin130 3) Law of cosines: cosB= ((10.7)^2 - (9.5)^ - (12.4)^2 )/[(9.5)(12....
Sunday, September 23, 2007 at 11:07am by bobpursley
physics
Draw the vector diagram. Draw the N vector. THen, from the top of the N vector, sketch the flow of water vector. Now, the resultant is from the bottom of the N vector to the tip of the water vector. Using the law of cosines: R^2=B^2 + W^2 -2BWcos60 and the direction, from the ...
Sunday, September 4, 2011 at 9:51am by bobpursley
Trig
sides of triangle are (2+2.5),(2+3),(2.5+3) Start with law of cosines c^2=a^2+b^2-2abCosC label the sides a,b, c solve for angle C Then, use the law of sines a/SinA=c/SinC solve for angle A then use the fact that the sum of the angles is 180 deg, find angle B. check angle B ...
Sunday, February 17, 2013 at 9:36pm by bobpursley
Trig
Should the triangle be solved beginning with Law of Sines of Law of Cosines. Then solve the triangle. Round to the nearest tenth. A=56 degrees, B=38 degrees, a=13. Sines. I get confused on the formula. I know C=86 degrees
Friday, March 7, 2008 at 1:38pm by Jon
physics
I don't know what you have been taught. If you make a sketch, put the vectors head to tail, you have a triangle with two known legs, with the included angle (64deg) known. You can find the other side with the law of cosines. Then, with the law of sines, find the angle of ...
Tuesday, September 21, 2010 at 6:37pm by bobpursley
Physics
You can use the law of cosines here, but it is much more complicated. Two sides a,b are the vectors (head to tail), and c is the unknown side. The angle between the vectors is 180-60=120 draw a triangle to prove that. c^2=a^2+b^2-2ab*cos120 = 2*tension^2( 1-cos120) =2 tension^...
Sunday, April 3, 2011 at 1:36pm by bobpursley
Pre Calculus/Trig
You can get c with the law of cosines. Then, draw the figure. You have a multitude of similar triangles. The idea is to use them to find the measure of radius, the distance on many of those triangles.
Thursday, October 11, 2007 at 9:04pm by bobpursley
trig
using the law of cosines, the distance d is d^2 = 16^2 + 18^2 - 2(16)(18)cos110° d = 27.87 ft
Sunday, April 21, 2013 at 2:45pm by Steve
pre calculus
I assume you have had trig. Draw the figure. if it has a 70 degree acute angle, the other angle is 110. Use the law of cosines: shortdiag^2=6^2+7^2-2*6*7Cos70 longdiag^2=6^2+7^2-2*6*7Cos110
Wednesday, December 1, 2010 at 7:37pm by bobpursley
Trig
Assuming that A is the side opposite angle A, etc. Use the law of cosines: B^2 = A^2 + C^2 - 2AC cosB = 31.6^2 + 42.8^2 - 2(31.6)(42.8)(-.7313) = 4808.53 B = 69.3
Tuesday, November 22, 2011 at 11:06am by Steve
math
Use cosines law for SSS and SAS use sine law for others a) cosine law b) sine law
Thursday, March 25, 2010 at 11:05am by Reiny
Geometry
How do I find the mesurements of a isosceles triangle? I have the sides set as E=72 degrees, and the legs that come down are 15, and i have to find D and F..I have no clue how to do this. Or any idea of a formula to help with this. The two base angles are equal, and the sum of...
Sunday, May 27, 2007 at 2:42pm by Fonda
geometry
not at all. See law of sines, law of cosines.
Monday, January 9, 2012 at 9:30pm by Steve
Geometry
Try using the Law of Signs or the Law of Cosines.
Sunday, September 5, 2010 at 7:27pm by Henry
Algebra 2
Using the information given about a triangle, which law must you use to solve the triangle? Law of Sines, Law of Cosines, or Neither. ASA SSS SAS AAA SSA AAS
Monday, February 13, 2012 at 12:19pm by Katy
Trig
Can you show us how to use law of cosines to show that theta equals inverse cosine of (a^2 + b^2 -484)/(2ab) where a^2=(7+xcos a)^2 + (28-xsin a)^2 and b^2=(7+xcos a)^2 + (xsin a -6)^2? Thanks
Tuesday, April 17, 2012 at 8:00am by Lisa
Trigonometry
Law of sines: a/SinA = b/SinB solve for B from that. Then solve for C knowing A and B, and the sum of angles is 180 Then, find c with the law of sines, or law of cosines.
Sunday, January 24, 2010 at 7:19pm by bobpursley
Trig - check my answers plz!
1. (P -15/17, -8/17) is found on the unit circle. Find sinΘ and cosΘ Work: P= (-15/17, -8/17) cosΘ = a value P = (a,b) sinΘ = b value Answer: cosΘ = -15/17 sinΘ = -8/17 2. Should the triangle be solved beginning with Law of...
Tuesday, February 3, 2009 at 4:27pm by Piper
trig
LAW OF COSINES: CosA=(b^2+c^2-a^2)/2bc Cos46=(16+64-a^2)/2*4*8. Cos46=(80-a^2)/64, 80-a^2=64Cos46, a^2=80-64Cos46=35.54, a=5.96=6.0 LAW of SINES: SinB/b=SinA/a SinB/4=Sin46/6,Multiply both sides by 4: SinB=4*Sin46/6=0.4791, B=28.66=28.7
Tuesday, June 29, 2010 at 3:59pm by Henry
Calculus
A plane flying with a constant speed of 14 km/min passes over a ground radar station at an altitude of 7 km and climbs at an angle of 45 degrees. At what rate, in km/min is the distance from the plane to the radar station increasing 2 minutes later? I know you use law of ...
Saturday, October 14, 2006 at 2:09am by Michael
geometry
If a triangle has sides of lengths a and b, which make a C-degree angle, then the length of the side opposite C is c, where c2 = a2 + b2 − 2ab cosC. This is the SAS version of the Law of Cosines. Explain the terminology. Derive an equivalent SSS version of the Law of...
Monday, April 12, 2010 at 12:12am by hodna
trig
123-90 = 33 south of east law of cosines b^2 = a^2 + c^2 - 2 a c cos B b^2 = 200^2 + 300^2 - 2(200*300) cos 33
Sunday, March 11, 2012 at 7:09pm by Damon
trigonometry
okay so in my review packet under the "calculator" section it says law of sines and law of cosines. i dont know what that means. i wikipedia-ed it but i dont understand what they are saying. Please help!
Thursday, January 22, 2009 at 8:50pm by Spencer
Physics
i was woundering if you could point in the right direction i'm looking for very easy physics topics that makes use of the law of sines or the law of cosines Thanks
Monday, October 19, 2009 at 6:14pm by Physics
college Trig
Triangle ACB, C = 70 deg, a = 70, b = 50 find side c Use Law of Cosines--Case III (Given two sides and the included angle) c^2 = a^2 + b^2 - 2ab cos C c^2 = 70^2 + 50^2 - 2(70)(50) cos 70 Solve for c
Sunday, February 13, 2011 at 5:58pm by helper
Pre-Calc Helppp!!!
Find the indicated angle è. (Use either the Law of Sines or the Law of Cosines, as appropriate. Assume a = 110 and c = 136 (angle B=38) Round your answer to two decimal places.)
Thursday, April 5, 2012 at 11:42pm by Mike
CALCULUS
If you want the magnitude of the resultant, use the law of cosines. Drawing a figure will help. The law of sines can get you the sine of any angle of the triangle formed by the two velocity vectors and the resultant. it's easier using components, but the answer will be the...
Saturday, June 28, 2008 at 3:31pm by drwls
Maths/Right Triangle
There must be a relationship (formula) between the lengths of the sides of a Right Triangle and the angles opposite these sides. Help Please. Mike. Look at the law os sines or cosines. Law of sines a/sin(A) = b/sin(B) = c/sin(C) Law of cosines a2 = b2 + c2 - 2bc cos(A) b2 = c2...
Saturday, October 21, 2006 at 2:09pm by Mike
Pre-Calculus-Trig
Use the Law of Cosines to solve triangle ABC: cosC = (a^2 + b^2 - c^2) / 2ab. cos38=((440)^2+(300)^2-c^2)/2*440*300, 0.7880 = (283600 - c^2) / 264000, Cross multiply: -c^2 + 283600 = 208032, -c^2 = 208032 - 283600, -c^2 = - 75568, c^2 = 75568, c = 275ft. = Distance from A to B.
Sunday, January 30, 2011 at 7:57pm by Henry
Trigonometry
law of cosines b^2 = a^2 + c^2- 2ac cos B
Monday, May 30, 2011 at 4:45pm by Damon
math
check out law of cosines
Sunday, February 6, 2011 at 9:55am by dave
math
Compare with what? Law of Cosines: c^2=a^2 + b^2 -2abCosineC
Friday, October 30, 2009 at 11:33am by bobpursley
Physics
draw the figure. In my head, if R is the resultant Then the law of cosines R^2=268^2+310^2-2*268*310Cos70 The angle can be found by the law of sines, but you need to draw the figure and see how that works, after you get the angle in the law of sines, you have to orient it with...
Tuesday, September 20, 2011 at 7:28pm by bobpursley
Math (Trig)
I don't think your equation is an identity, I tried several angles and the Left Side is not equal to the Right Side. Check your typing. In general, I try to prove these type of identities by changing everything to sines and cosines, unless I can recognize one of the common...
Wednesday, April 9, 2008 at 2:06pm by Reiny
math
solve each triangle using either the Law of Sines or the Law of Cosines. If no triangle exists, write “no solution.” Round your answers to the nearest tenth. A = 23°, B = 55°, b = 9 A = 18°, a = 25, b = 18
Friday, December 9, 2011 at 11:40pm by Rick
MATH
Using the law of cosines: x1^2 = 100^2 + 500^2 - 2*100*500*cos(85) x1^2 = 100^2 + 500^2 - 2*100*500*0.087155743 x1^2 = 251284.4257 x1 = 501.28 Similarly, for the other side, we know that the angle between the tower and the ground is 90+5=95. Using the law of cosines to solve ...
Tuesday, February 16, 2010 at 11:31pm by PATEL
Summer School Calculus
I would break the two vectors into components N, and E. For instance, the N component of the first is 7cos30. Then, after finding the components, add the N, then the N. Then use trig to find the resultant. Another way is to use the law of cosines and law of sines, but I dont ...
Saturday, June 28, 2008 at 8:31pm by bobpursley
trig
Express sin4xcos3x as a sum or differences of sines and cosines
Thursday, February 21, 2013 at 4:04pm by Fabian onyango ndege
math
Please look up "law of cosines" with Google.
Sunday, August 21, 2011 at 5:39pm by Damon
physics (vector)
Use the law of cosines, and solve for the cosine.
Monday, June 6, 2011 at 11:38pm by drwls
math(advanced algebra&trigonometry)
Law of cosines c^2 = a^2+ b^2 - 2 a b cos C
Saturday, April 10, 2010 at 1:56pm by Damon
Trigonometry
When using the Law of Cosines, how do I know if two triangles exist?
Tuesday, January 19, 2010 at 6:14pm by Laurianne
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 7:35am by Reiny
URGENT SUMMER SCHOOL CALCULUS
Law of cosines: The side you are looking for is opposite the 30 degree angle. c^2=15^2+8^2-2*15*8cos30 I get a magnitude of about 9 For the angle, Iwould use law of sines. Draw a sketch first.
Sunday, June 29, 2008 at 8:30pm by bobpursley
Trig
I n my previous answer, I solved it using a vector method, not a trig method. Both will give the same answer. To use trigonometry, draw a triangle with two sides representing the two distances and directions of the planes after three hours. The angle between these two sides is...
Saturday, May 17, 2008 at 12:40pm by drwls
math
Correction: Used Law of cosines to find one of the angles.
Thursday, July 26, 2012 at 10:20pm by Henry
trig
I would say they are the definitions of tan, cot, sec and csc in terms of sin and cos, tan = sin/cos cot = cos/sin csc = 1/sin sec = 1/cos sin^2 + cos^2 = 1 the law of sines the law of cosines sin (a + b) = sin a cos b + cos b sin a cos (a + b) = cos a cos b - sina sin b tan (...
Monday, April 7, 2008 at 12:20am by drwls
trig
values of cosines lie between -1 and +1 so minimum value of 2cosx + 2cosy = 2(-1) + 2(-1) = -4 when both x and y = 3π/2
Wednesday, April 17, 2013 at 8:27am by Reiny
precal 2
use the law of cosines to solve the triangle a=55, c=72, b=25
Sunday, July 31, 2011 at 5:16pm by raven
precal 2
use the law of cosines to solve the triangle A=55 degree, c=10, b=3
Friday, July 29, 2011 at 4:05pm by danielle
physics
draw the figure, label the angles in the triangle, and use the law of cosines.
Thursday, September 25, 2008 at 3:55pm by bobpursley
trig
Ship A went 30 * 3 = 90 km Ship B went 45 * 3 = 135 km So we have a triangle with legs of 90 and 135 and an opposite side of 120 Use the law of cosines 120^2 = 90^2 + 135^2 - 2*90*135 cos T solve for T
Sunday, February 24, 2008 at 9:16pm by Damon
Trig.
I have answers for these problems, but I wanted to check if I had them right because I wasnt sure on some of them....Thanks. Solve the triangle: 1. a=4, b=8, alpha=30 deg. answer- beta=90 deg., gamma=60 deg., c=7 2. a=5, b=7, alpha=30 deg. answer- beta=44 deg., gamma=106 deg...
Tuesday, December 12, 2006 at 6:00am by ashley
geometry
call the vertices A,B,C Use law of cosines to find A. area = 1/2 * AB * AC*sinA
Thursday, February 16, 2012 at 10:56pm by Steve
Math
Use the law of cosines to find the angle Q between the given vectors. v = 3i + j w = 2i - j
Tuesday, March 30, 2010 at 10:35pm by Hannah
calculus
You have two angles in the triangle (draw the figure), and one side. With two angles, you can figure the last angle. I would use the law of sines, but you could use the law of cosines.
Wednesday, May 9, 2012 at 11:10am by bobpursley
trig 26
simplify to a constant or trig func. 1. sec ²u-tan ²u/cos ²v+sin ²v change expression to only sines and cosines. then to a basic trig function. 2. sin(theta) - tan(theta)*cos(theta)+ cos(pi/2 - theta) 3. (sec y - tan y)(sec y + tan y)/ sec y combine ...
Sunday, March 14, 2010 at 1:25pm by hilde
trig/math
3. Draw vectors u and v tail-to-tail so that they make a è-degree angle. Draw the vector u − v, the third side of the triangle, and check to see that it points in the right direction. (a) Solve for cos è using the SSS version of the Law of Cosines, ...
Sunday, April 18, 2010 at 7:36pm by sarah
Math
Law of Cosines C^2=a^2+b^2-2abCosC so label your sides a, b, c you want angle c to equal RAB so RB=C RA=a AB=b
Saturday, March 3, 2012 at 4:25pm by bobpursley
Pre-Cal
Use the law of cosines to find the angle of Q between the given vectors. v=3i + j w=2i - j
Sunday, March 28, 2010 at 5:08pm by Hannah
Vectors
Use the law of cosines: You know two sides, and the included angle. Draw the figure before you compute.
Sunday, June 15, 2008 at 2:54pm by bobpursley
Trig
rewrite all in terms of sines and cosines, find a common denominator and "simplify". BTW, I got it as far as cotA(cosA - 1)
Wednesday, April 16, 2008 at 9:58pm by Reiny
Trigonometry
first you would want to convert feet to miles or vice versa. Then, youll want to build the triangle because you start at 40,000 feet and then that goes at a six degree angle along the x axis for 20 miles. So you will want to put this on a cartesian plane because from there you...
Sunday, March 11, 2012 at 11:43pm by john
pre-Cal(Please help)
Use the law of cosines to find the angle Q between the given vectors. v = 3i + j w = 2i - j
Wednesday, March 31, 2010 at 5:34pm by Hannah
Law of cosines
A tree on a hillside casts a shadow 215 ft down the hill. If the angle of inclination of the hillside is 22 degree to the horizontal and the angle of elevation of the sun is 52 degreess, find the height of the tree. TIA sorry this is law of sines. sorry for the confusion.
Tuesday, March 27, 2007 at 4:36pm by Jen
Calc
Not a right triangle? You know angle, or SAS. you want the other distance (S) I think I would start with the law of cosines: c^2=a^2+b^2 -2ab CosC
Sunday, October 31, 2010 at 11:42am by bobpursley
geometry
I am studying for my GRE exam into grad school and it's been a very long time since I've done geometry. I have a problem to which I need to solve for the area of a triangle but I do not have the base. I do have all angles but I cannot remember how to convert angles ...
Monday, July 10, 2006 at 7:29pm by Kaytee
physics
This is pretty simple trig and SNells law, along with some geometry. Draw the figure, and figure the intercept point at the surface, then snells law, then use trig to find the intercept at the bottom. You willhave to decide what the angle of incidence is: is it 24 deg, or 66 ...
Saturday, September 10, 2011 at 8:28pm by bobpursley
physics
You have already said what the magnitudes of A and B are. Call them |A| and |B|. I assume you want the magnitude of vector sum A + B Use the law of cosines. |A+B|^2 = |A|^2 + |B|^2 - 2|A||B| cos 60
Wednesday, September 8, 2010 at 9:15pm by drwls
math
a) use law of cosines b) use hero's formula
Friday, November 25, 2011 at 11:02am by John
Agrebrla
the angle opposite the biggest side is biggest. law of cosines 5^2 = 1^2+3^2 - 2 *1*3 cos A
Friday, December 2, 2011 at 5:21pm by Damon
law of cosines and sines
m<C=70,c=8,m<30 solve the triangle
Thursday, April 28, 2011 at 10:34pm by Anonymous
Physics
Numbers out require numbers in. Add components or use the law of cosines.
Monday, August 20, 2012 at 11:20pm by drwls
physics
It is the vector sum of the two forcesw. Note that they are 135 degrees apart. You can either add x- and y- components (separately) or use the law of cosines.
Tuesday, April 3, 2012 at 11:56pm by drwls
MATH
a web search for "law of cosines for vectors" shows various articles. The 4th entry at tri-c dot edu ha a diagram and explanation.
Wednesday, November 2, 2011 at 9:37pm by Steve
Math
Use the law of cosines to find side a. a^2=b^2+c^2 -2bcCosA Then area= sqrt [s(s-a)(s-b)(s-c)] where s is half the perimeter. There are about 100 other ways to work this.
Sunday, February 20, 2011 at 6:00pm by bobpursley
Calculus
Thanks. I didn't think my answer made any sense. I tried to do too much, instead of taking the derivative of the law of cosines equation, which I had used initially...
Sunday, November 16, 2008 at 4:34pm by Joe
geometry
find the third side with the law of cosines. Then, area= sqrt((s-a)(s-b)(s-c)s) where s is 1/2 the perimeter. Perimeter then is a+b+thirdside.
Sunday, April 11, 2010 at 4:57pm by bobpursley
math 10
sines, cosines, and tangents As you said before, you are in grade 11 Math. If you don't know what these trig ratios are, I think you are in above your head.
Thursday, February 4, 2010 at 5:01pm by Reiny
math
how is 2 not A? my book says to solve an oblique triangle: two angles and any side, and two sides and an angle opposite one of them is solved by Law of Sines. Two angles and their included angle or three sides are solved by Law of Cosines.
Monday, February 11, 2008 at 2:38pm by Jon
geometry(triangle)
There's probably a good geometric way to do it, but here's a trig way, using the law of cosines. AB^2 = 6^2+15^2 - 2 * 6 * 15 * (-1/2) = 351 AC^2 = PC^2 + 15^2 - 2*PC*15(-1/2) BC^2 = PC^2 + 6^2 - 2*PC*6(-1/2) AC^2-BC^2 = AB^2 = 351, so 351 = 189 + 9PC 9PC = 162
Monday, February 25, 2013 at 11:31am by Steve
math
The Law of Cosines: CosB = (-b^2 + c^2 + d^2)/2cd. cosB=(-1681 + 4225 + 2704)/6760=0.77633 B== 39.1o
Tuesday, April 16, 2013 at 5:27pm by Henry
Nautical/Maths
I suppose I should explain more about the haversines (half angle sines) The log of a negative number is undefined. Remember the definition of base ten log: 10^log x = x Now ten to what power is a negative number? No such thing/ BUT, here we may very well have sines and cosines...
Friday, January 4, 2008 at 3:49am by Damon
physics
What is the meaning of the / in front of 3.0 ? Why don't you just write the phase shift angle in radians? You can get the angle by adding the phasors, using the law of cosines
Friday, December 24, 2010 at 10:38am by drwls
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