Wednesday

April 16, 2014

April 16, 2014

Number of results: 222,799

**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*

**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*

**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**

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*

**Trigonometry**

You'll have to use the Ambiguous Case for the Law of Sines. I would give you a few links to some good websites. However, I'm not allowed. Try googling "ambiguous case law sines" without the quotes for some relevant information.
*Monday, November 26, 2007 at 10:40am by Michael*

**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*

**math**

I drew the figure. I saw two triangles. In one, I used the law of sines (I knew two angles, one side) to find the other side, then used that side (common to the other triangle, to solve height with the law of sines again.
*Sunday, August 21, 2011 at 4:13pm by bobpursley*

**Math**

1. use the law of Sines to solve (if possible) the triangle for the value of c. A=27.5degrees, a=15cm, b=36.4cm 2. use the law of Sines to solve (if possible) the triangle for the value of c. B=52degrees, a=8cm, b=12cm How do I solve these type of problems?
*Tuesday, June 4, 2013 at 2:17pm by mysterychicken*

**bearing**

Ok, the angle at p is between 310 and 075,or an angle of 125 (check that). YOu have two sides, one angle find another angle with the law of sines. then find the last angle (sum of angles is 180) then find the last side by law of sines. Make certain you draw the figure, life is...
*Thursday, January 7, 2010 at 3:32pm by bobpursley*

**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*

**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*

**Maths**

you give |<QPR|= 43°, then ask for |<QPR|. assuming you want to know the rest of the values for the triangle, use the law of sines: sin43°/5 = sinR/7 R = 72.7° so, angle Q = 180-(43+72.7) = 64.3° all we have left is side q, so back to the law of sines: q/sin64.3° = 5/...
*Thursday, October 18, 2012 at 2:53pm by Steve*

**Math repost - girly93**

The law of cosines, in one form, is a^2 = b^2 + c^2 - 2bc cos A The law of sines is a/sin A = b/sinB = c/sinC = 2R Your cosA equation is incorrect. (1/2) a b sin C is a formula for the area of a triangle, and has nothing to do with the law of sines. For more about these ...
*Monday, February 9, 2009 at 7:01am by drwls*

**geometry**

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

**math HELP!**

Since angle B is 60 degrees, and A is less (30 degrees) the law of sines says that the length of side a is less than that of side b. The length of side c is given by the lasw of sines: c/sin 90 = c = b/sin60 c = 4.0/0.866 = 4.619 km
*Wednesday, January 9, 2013 at 11:58pm by drwls*

**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*

**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*

**Math**

Then use the law of sines again.
*Tuesday, June 4, 2013 at 2:17pm by Steve*

**the law of sines/math**

Find the area of each figure ??
*Monday, April 9, 2007 at 12:59pm by Sabrina*

**Math**

answered elsewhere. Another law of sines problem
*Saturday, May 17, 2008 at 12:23pm by drwls*

**Math**

apply the law of sines to get B or A. recall that A+B+C = 180°
*Tuesday, June 4, 2013 at 2:17pm by Steve*

**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**

You know angle B as 90-A. Law of sines: c/SinC=b/SinB solve for b.
*Friday, February 10, 2012 at 7:19pm by bobpursley*

**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*

**Math**

I would use the law of sines to find angle B. SineB/b=SinA/a
*Thursday, October 2, 2008 at 9:30pm by bobpursley*

**math**

This is a brilliant qn so i will only give you a hint: Use the law of sines.
*Monday, June 17, 2013 at 2:53pm by exactly*

**trig**

I see in the triangle the following: obtuse angle: 180-63.8, opposite side 106ft, another side 64. Law of sines to get the guy angle: 64/sinA=106/sin(180-63.8) solve for angle A. Then, knowing angle A, you can find the third angle at the top between the guywire and the ...
*Thursday, February 18, 2010 at 10:49am by bobpursley*

**math**

list two congruence conditions for the law of sines yields a unique solution
*Tuesday, April 27, 2010 at 11:11am by carol*

**math - trig**

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

**Trig**

Use the law of sines to solve for <C. (sin <C)/c = (sin <B)/b Use <A + <B + <C = 180 to solve for <A Uze the law of sines to solve for a. The triangle area is (1/2) c sin B * a
*Wednesday, November 4, 2009 at 6:07pm by drwls*

**Math - Trig**

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

**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*

**Physics**

Given the three sides (the largest must be less than the sum of the other two) Find any angle from cosA = (b^2 + c^2 - a^2)/2bc. The remaining angles can be derived from the Law of Sines. Alternatively, find any angle, A for instance, using tan(A/2) = r/(s - a) where s = (a + ...
*Sunday, May 6, 2012 at 11:25am by tchrwill*

** 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**

you can save some work by recognizing that the smallest angle is opposite the shortest side (law of sines)
*Wednesday, October 17, 2012 at 2:03pm by Steve*

**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*

**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*

**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*

**Trigonometry**

a^2=b^2+c^2-2bcCosA CosA= 1/2 * (b^2+c^2-a^2)/bc A= 139 deg, you are right on that. Lets find angle B, by the law of sines sinB/b=SinA/a or sinB= .3269 or B=19 deg Lets find angle C by the law of sines SinC/c=SinA/a or SinC=.3718 C=21.8 deg check: do the sum of angles = 180 ...
*Tuesday, January 19, 2010 at 6:25pm by bobpursley*

**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*

**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*

**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*

**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*

**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*

**trig**

can't understand the question. it's law of sines. what's the S,a,b, etc ?
*Friday, December 2, 2011 at 11:18am by Anonymous*

**trig**

law of sines: b/SinB=a/SinA
*Wednesday, January 25, 2012 at 12:12pm by bobpursley*

**Trig - Law of sines and cosines**

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

**Math**

Use the Law of Sines to solve triangle ABC when a=1.43,b=4.21,and A=30.4 degrees. If no solution exists, rite none.
*Tuesday, June 9, 2009 at 10:35am by Sally*

**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*

**math**

by the law of sines, the side opposite the larger angle has the larger length.
*Friday, January 24, 2014 at 2:24pm 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*

**Math**

draw the figure. Then, law of sines. sinNBA/120 =sinBAN/distance solve for distance.
*Sunday, August 21, 2011 at 4:10pm by bobpursley*

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

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

**Math**

a. a neat way to get area is this relation area=sqrt(s(s-a)(s-b)(s-c)) where s is half the perimeter, and a,b,c are the sides. You can get the last side with the law of sines.
*Friday, June 7, 2013 at 11:20am by bobpursley*

**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*

**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 into their ...
*Monday, July 10, 2006 at 7:29pm by Kaytee*

**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*

**precal 2 **

use law of sines to solve the triangle A=58 degree, a=4.5, b=12.8
*Sunday, July 31, 2011 at 5:12pm by raven *

**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*

**Math**

No, you have done the law of sines wrong> SinA/a=SinB/b You have the sides mixed up. Sin63/R=Sin45/200 check my thinking.
*Saturday, May 17, 2008 at 10:39pm by bobpursley*

**Maths**

using the law of sines, PF/sin43° = 490/sin36° plug and chug
*Monday, April 15, 2013 at 4:17pm by Steve*

**Geometry**

You should rethink 4. 2. is 'sometimes'. Consider the isosceles case. 3. is true because of the law of sines
*Tuesday, May 28, 2013 at 9:02pm by drwls*

**Maths/Law of sines**

I have a spherical triangle and I know 1 angle 31.3 degrees and all 3 sides which are 1624, 2118.4 and 1078.85 nautical miles. In order to find the other 2 angles I know I must use the law of sines: sin A over a = sin B over b = sin C over c If angle A is 31.3 degrees and side...
*Sunday, January 13, 2008 at 1:18pm by Mike*

**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**

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**

I was just wondering if anyone can explain the Law of Sines- "Ambiguity Case" to me, in a way that is easy to understand.
*Saturday, November 15, 2008 at 2:15pm by Ariel*

**pre cal 2**

use law of sines t solve the triangle A=5degree 40', B=8degree 15', b=4.8
*Friday, July 29, 2011 at 4:01pm by danielle*

**trig**

since A+B+C = 180, we can see that C = 78° Now, use the law of sines: c/sinC = a/sinA
*Friday, February 22, 2013 at 5:49pm by Steve*

**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*

**Math**

Use the law of sines. 25/sin 41 = (hypotenuse)/sin 90 = hypotenuse = x x = 38.1
*Monday, May 11, 2009 at 9:12pm by drwls*

**Calculus**

Use the Law of sines to find the remaining sides. 1. a=40 degrees; Beta=87; c=115
*Wednesday, July 20, 2011 at 9:30pm by Josh *

**Physics**

For the Law of Sines state the formulas, explain them in your own words, and draw a diagram illustrating one of them.
*Sunday, September 11, 2011 at 12:01am by Lena*

**trig**

Use the Law of Sines to solve each triangle. P=110 degrees p=125, r=200 in triangel PQR
*Saturday, November 26, 2011 at 10:56pm by Tyler*

**Help In Math!**

well, first off you know that since a<c<b, A<C<B (law of sines) That means (b)
*Tuesday, May 21, 2013 at 3:38pm by Steve*

**Math**

The law of sines is the way to do this. That is your method #2. Method #1 only applies to right triangles. You made some algebra errors, however. sin B = 0.4589 B = 27.3 degrees 180 - B - C = A , not what you wrote.
*Saturday, May 24, 2008 at 4:26pm by drwls*

**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*

**Trigonometry**

by law of sines, 7.6/sin38 = 9.2/sinS knowing P+S+Q = 180 degrees, now you know Q all information about R is irrelevant
*Saturday, September 22, 2012 at 5:58pm by Steve*

**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*

**Maths/Law of sines**

You must use the fact that one degree around the earth is 60 nautical miles to get distances as angles.
*Sunday, January 13, 2008 at 1:18pm by Damon*

**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*

**Math**

sin B = 0.8*sin 35 I used the law of sines, which you wrote correctly but did not calculate correctly, to get sin B. Then I took the arcsin of that to get the angle B itself
*Saturday, May 24, 2008 at 6:38pm by drwls*

**law of cosines and sines**

m<C=70,c=8,m<30 solve the triangle
*Thursday, April 28, 2011 at 10:34pm by Anonymous*

**Math**

The third angle of the triangle is PRT = 180-63-72 = 45 degrees. The side with length 200 m is opposite this angle. Use the law of sines for the other side lengths.
*Saturday, May 17, 2008 at 10:39pm by drwls*

**Math**

There also can be used the law of sines... c/Sin90=AC/sinB sinB= AC/(sqrt(AC^2+BC^2) and B= arc sin ( ) I am not certain what the purpose of your assignment is.
*Friday, June 19, 2009 at 9:51am by bobpursley*

**Physics please help me!**

Prove the law of sines for triangle ABC i.e sinA/a=sinB/b=sinC/c.0 please help me!!!
*Wednesday, November 16, 2011 at 9:58am by Meron*

**physics**

I dont know where theta 1 and theta 2 are, but have you considered drawing the figure, then using the law of sines?
*Sunday, January 26, 2014 at 2:11pm by bobpursley*

**Calculus**

Using Law of Sines: Need Help 1. alpha=70 Degrees Beta=38 Degrees a=35
*Thursday, October 13, 2011 at 12:08am by Josh*

**Trigonometry**

Find the 3 angles of the triangle, then use law of sines to find a, the side opposite angle A then, d = a sin34°
*Monday, February 4, 2013 at 10:24pm by Steve*

**cmsd**

Then each of the acute angles is 40, and the the angle bisected by the shorter diagonol divides into two 70deg law of sines: 50mm/Sin40= s/Sne70
*Wednesday, February 20, 2013 at 7:42pm by bobpursley*

**Physics**

Draw the figure. To get distance, use Pythagoras relationship. To get displacement, add the direction. Use the law of sines to get the angle.
*Wednesday, September 7, 2011 at 4:40pm by bobpursley*

**go with kuai**

kuai's solution is better, because the assignment is evidently involving the law of sines/cosines. also, less work!
*Friday, November 8, 2013 at 11:56pm by Steve*

**Math(Please help)**

Use the law of sines first sin C/c = sin A/a that gives you A then B = 180 - A - C then use sin B/b =sin A/a
*Saturday, March 27, 2010 at 9:04pm by Damon*

**trigonometry (repost) Mathmate**

You can check your value of c by using the Law of Cosines, with the original calyues of a, b and C. c^2 = a^2 + b^2 -2ab cos C c^2 = 460.32 +2141.84 -1986.34*0.8436 c = 30.44 That is not what you came up with. There appears to be something wrong with your initial angles. The ...
*Saturday, February 26, 2011 at 3:37am by drwls*

**maths**

Choose three options which are true: a) an angle of 150 degrees is equivalent to 2pie/3 radians. b) Cos 0 = cos (0 – pie/2) for al values of 0. c) Sin 0 = cos (0 – pie/2) for all values of 0. d) If triangle ABC has a right angle at B, then sin A = cos C e) In any triangle ABC...
*Friday, June 8, 2007 at 6:31pm by kat*

**Math**

The law of sines says that the ratio of the common bisector line to the sine of the unbisected side angles must be the same. Therefore the two side angles must be equal. That makes the triangle isosceles.
*Saturday, August 20, 2011 at 7:37am by drwls*

**Math**

I assume these ropes are on oppsite sides of the Balloon. If so, you have two sides, one angle. Use the law of sines to get the other ground angle. sin65/115 =sinTheta/120 solve for sinTheta, then Theta=arcsin.. Now, since you have both ground angles, the rest is easy. You ...
*Thursday, June 6, 2013 at 9:08pm by bobpursley*

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