Wednesday

April 16, 2014

April 16, 2014

Number of results: 16,545

**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't know that, ...
*Tuesday, October 16, 2012 at 6:14pm by Steve*

**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 by McKenna Louise*

**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 Add ...
*Tuesday, December 12, 2006 at 5:45pm by ashley*

**Trig - Law of sines and cosines**

ABC is an equilateral triangle with side length 4. M is the midpoint of BC, and AM is a diagonal of square ALMN. Find the area of the region common to both ABC and ALMN. I drew the diagram but I don't know how to find the answer? I think it has something to do with the law of ...
*Tuesday, November 19, 2013 at 8:09pm by Sam*

**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 - Law of sines and cosines**

Thanks a lot
*Tuesday, November 19, 2013 at 8:09pm by Sam*

**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*

**Trig**

in each case, use law of cosines to get c or C: c^2 = a^2+b^2-2ab cosC Now, having all 3 sides, and angle C, use law of sines to get A and B: sinA/a = sinB/b = sinC/c
*Wednesday, January 15, 2014 at 1:01am by Steve*

**Trig-Medians and law of cosines and sines**

thanks a lot man!
*Wednesday, November 20, 2013 at 11:53am by Sam*

**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 - 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 by McKenna Louise*

**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)(12.4)] ...
*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*

**Trig-Medians and law of cosines and sines**

In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is BC? Not making sense to me, I think the answer must be simple, but I don't know how to solve I applied the law of sines but to no avail. Help is appreciated, thanks.
*Wednesday, November 20, 2013 at 11:53am by Sam*

**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 it (...
*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*

**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 8:34pm by Steve*

**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*

**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 1:59pm by Steve*

**Precalculus**

I know I posted this before,but I did not learn it your way .....instead I drew a picture and use the law of cosines....is there a way you can show me the work using law of cosines?? A plane flies 30 mi on a bearing of 200 degrees and then turns and flies 40 mi on a bearing of...
*Thursday, December 12, 2013 at 9:31pm by Anonymous*

**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**

Try using the Law of Signs or the Law of Cosines.
*Sunday, September 5, 2010 at 7:27pm by Henry*

**geometry**

not at all. See law of sines, law of cosines.
*Monday, January 9, 2012 at 9:30pm by Steve*

**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*

**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*

**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**

by law of cosines, c^2 = a^2 + b^2 - 2ab cosC Now, having a,b,c and C, use the law of since to get A,B. sinA/a = sinB/b = sinC/c
*Tuesday, March 18, 2014 at 9:40am by Steve*

**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 Sines or Law of Cosines...
*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*

**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*

**Trigonometry/Geometry - Law of sines and cosines**

In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles ABC and DEF such that AB = DE, BC = EF, and angle A = angle D, then we cannot deduce that ABC and DEF are congruent. However, there are a few special cases in which...
*Thursday, November 21, 2013 at 4:03pm by Sam*

**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*

**math**

Thank you steve, but something must be wrong with this problem. We have not learned law of cosines or law of sines yet and the correct answer is ad?? Baffled at how to get this based on what we are studying, SAS and SSS inequality theorems.
*Friday, January 24, 2014 at 2:24pm by Jacob*

**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*

**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*

**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 same.
*Saturday, June 28, 2008 at 3:31pm by drwls*

**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*

**Pre calculus**

Why is it dangerous to use the law of sines to find an angle but is not dangerous to use the law of cosines? Please explain this thoroughly.
*Sunday, December 8, 2013 at 10:00pm by Anonymous*

**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*

**Trignometry**

use the law of cosines to start and find side c. Then use the law of sines to find remaining angles.
*Sunday, June 30, 2013 at 1:25pm by Joe*

**math**

Compare with what? Law of Cosines: c^2=a^2 + b^2 -2abCosineC
*Friday, October 30, 2009 at 11:33am by bobpursley*

**math**

check out law of cosines
*Sunday, February 6, 2011 at 9:55am by dave*

**Trigonometry**

law of cosines b^2 = a^2 + c^2- 2ac cos B
*Monday, May 30, 2011 at 4:45pm by Damon*

**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*

**trig**

I usually change all I can to sines and cosines. Have you tried that ?
*Thursday, January 16, 2014 at 11:54pm 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*

**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*

**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*

**trig**

Express sin4xcos3x as a sum or differences of sines and cosines
*Thursday, February 21, 2013 at 4:04pm by Fabian onyango ndege*

**Trigonometry**

When using the Law of Cosines, how do I know if two triangles exist?
*Tuesday, January 19, 2010 at 6:14pm by Laurianne*

**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*

**physics (vector)**

Use the law of cosines, and solve for the cosine.
*Monday, June 6, 2011 at 11:38pm by drwls*

**math**

Please look up "law of cosines" with Google.
*Sunday, August 21, 2011 at 5:39pm by Damon*

**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*

**math**

The law of cosines can't be used when which of the following combinations of information is given?
*Monday, June 24, 2013 at 7:18pm by henna*

**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-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'C'. If ...
*Sunday, November 24, 2013 at 11:01pm by Sam*

**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*

**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*

**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*

**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 *

**Math**

Given the points A(0,0), B(4, 3) and C(2,9), what is the measure of angle ABC? Using Law of Cosines.
*Sunday, July 21, 2013 at 12:28am by jasmine*

**Trig-Medians and law of cosines and sines**

median length x = BC length median hits BC at D triangle ABC 4^2 = 3^2 + x^2 - 6 x cos B triangle ABD x^2 = 3^2 + (x/2)^2 - 3 x cos B ================================= 16 = 9 + x^2 - 6x cos B x^2 = 9 +x^2/4 -3 x cos B x^2 -6 x cos B -7 = 0 3x^2/4 +3 x cos B - 9 = 0 x^2 - 6 x ...
*Wednesday, November 20, 2013 at 11:53am by Damon*

**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*

**math**

by law of sines, bd/sin91 = 9/sin43 bd = 13.19 by law of cosines, ab^2 = 9^2 + bd^2 - 2(9)(bd)cos44 = 81 + 174 - 170.84 ab = 9.17 Looks like ab > ad
*Friday, January 24, 2014 at 2:24pm 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*

**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*

**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 fractions and simplify to a...
*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, expressing all lengths...
*Sunday, April 18, 2010 at 7:36pm by sarah*

**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*

**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*

**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*

**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*

**Trigonometry**

Using the Law of Cosines, how can I show that the measure of each angle of an equilateral triangle is 60 degrees?
*Monday, February 10, 2014 at 8:36pm by McKenna Louise*

**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*

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