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Topic: **Math homework calculator****Question:**
So, I need to use the Law of Cosines to solve the triangle given sides a, b, and c (55, 25, 72)
I understand that I'm gonna solve for the angles but then I need to turn my answer into a degree.
For example:
For cosA I got that it equals about 0.78. In the examples on the lesson page (this is an flvs course) they turn those answers into degrees but I don't know how and it doesn't talk about that so can someone explain to me how im supposed to do that? I don't have any calculators by the way, scientific, fraction, graphing, nothing lol..
please please?
:)
why is no one answering.....

July 19, 2019 / By Aubrie

convert radians to degrees using the ratio D/180 = R/pi In the case of 0.78 you would write D/180 = 0.78/pi The law of cosines says that a^2 = b^2 + c^2 - 2(ab)cos(a) In your problem you would write 55^2 = 25^2 + 72^2 - 2(25)(72)cos(0.78) If your instructor wants you to use degrees instead of radians, you should convert the 0.78 first. Be sure to use as many decimals as your calculator shows to avoid rounding errors. If you do not have a calculator, you can google "on-line calculator" and you should be able to use one on line. You will be unable to do the problem without one - unless you have an old trig table.

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seems like previous person already answered. if u still need help understanding what wikipedia said, i will help you. First, click the link u see above or type law of cosines on google and go to wikipedia. Then click [3.1 Using trigonometry] on the contents. You need the figure 4. to understand it. i will use A for alpha(the a shape) , and B for Beta(B shape), and C for gamma ( flipped r shape) since i cant type them. The equation you see in the explanation would be c=a cos (B) + b cos (A) a cos (B) represents the longer part of side c after the height separated it in to two parts. It is written as side a times cosine of angle B because cosine of angle B is equal to adjacent side of angle B (which is a cos (B)) over hypotenuse (which is side a) by using trig. ratio. so it would be written as acos(B)/a. Since you want only the adjacent side, u multiply side a to cos(B), which gives u acos(B) again. (Right now, u dont need to find what acos(B) is because it's a proof not solving equation. Just understand where it is origin from.) b cos (A) is same idea. This would turn in to c^2=ac cos (B) + bc cos (A)The purpose of this equation is that it is used for substitution to other equation you get later. Next, you multiply this equation by c because u want the c to be squared (c^2) so u can used this for pythagorian theorem later. It gives you c^2=ac cos (B) + bc cos (A) The next two equations (a^2= ac cos (B) + ab cos (C) and b^2=bc cos (A) + ab cos (C)) are expressed in the same way for different sides as the first equation. Then you use these functions as pythagorian theorem (a^2 + b^2 = c^2) a^2 + b^2 = ac cos (B) +ab cos (C) + bc cos (A) + ab cos (C) You need to put in order for right side so u can see whats going on. Since there are two of ab cos (C), put them first, then put other two (doesnt have to be order) It would be look like a^2 + b^2 = ab cos (C) + ab cos (C) + ac cos (B) + bc cos (A) (the equation on Wikipedia looks little bit different (the order is different) but it doesnt make difference) you can simplify this to a^2 + b^2 = 2 ab cos (C) + ac cos (B) + bc cos (A) Since ac cos (B) + bc cos (A) is equal to c^2 (This is when you substitute the equation). Finally, the equation look like this a^2+ b^2 = 2 ab cos (C) + c^2 After you move 2 ab cos (C) to the left side it looks like this a^2+ b^2 - 2ab cos (C) = c^2 which is same as the law of cosine formula c^2 = a^2+ b^2 - 2ab cos (C)

nicely if cosine starts off with a C and pythagorean starts off with a P, then all you should do is opposite the mollecular compound of the quantum fission formula, after which you should finally end up with proving the pythagorean and cosine AND the photonic fix theorem.

think of of the pendulum all a thank you to the appropriate and draw a genuine triangle representing the appropriate 0.5 of the swing. The hypotenuse is 80 cm, the backside and the horizontal leg is 6 cm (0.5 of the width of the swing) and you are able to desire to locate the top. Now set up your equation as Sin(theta) = opposite / hypotenuse. So theta = arcsin(6 / 80). in case you prefer to comprehend all 3 facets, you are able to desire to apply the Pythagorean theorem to verify the vertical section. then you definately could desire to apply any trig function.

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1. Given: mB = 70° mC = 15° a=12 solution: mA = 180° - (70° + 15°) mA = 180° - (85°) mA = 95° (b/sinB) = (c/sinC) = (a/sinA) --- Law of Sines, now take two at a time.. pair up each unknown side to the given side to avoid small deviations.. b/sinB = a/sinA b = ((12)(sin70))/(sin95) b = 11.3 units c/sinC = a/sinA c = ((a)(sinC))/(sinA) c = ((12)(sin15))/(sin95) c = 3.11 units. Basically you just use this: (b/sinB) = (c/sinC) = (a/sinA) in solving the problems, only take two at a time, and remember the sum of the angles of a triangle are 180, sorry, but too lazy to compute the other 6 questions.. :D in problem #8 , just draw the figure and label them accordingly, never ever forget to label, after solving the other 6, try it yourself and apply still the law of sines.. to solve 8, let x = distance of ship from 2nd station.. x/sin37 = 43/sin(180-(37+113)) x/sin37 = 43/sin(30) x = 43sin37/sin30 x = 51.8 km.

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