hypebeast Posted October 7, 2014 Report Share Posted October 7, 2014 I need to do this trigonometry excercise but I dont understand shit. Someone please help? Ok so we have a right triangle, its hypotenuseis 2√3. Its cosine is also 2/3. I have to determine every other function. Can anyone figure this out? What I've done by now is: cos= ? * √3 ? √3 ? √3 ? √3 ____ ___ = _______ = ________ = _______ 2√3 √3 2.3 6 3 ^rationalizing ^simplifying as you can see I found a way for my denominator to be 3, but I have no Idea how the numerator can be 2. Link to comment Share on other sites More sharing options...
Catman25 Posted October 7, 2014 Report Share Posted October 7, 2014 If h=2sqrt(3) and cos = 2/3, then adj/h = 2/3, or adj = 2/3*2sqrt(3) = 4sqrt(3)/3. Using pythagorean, the last side is sqrt(20/3). So use those to find everything. Link to comment Share on other sites More sharing options...
Nathanael D. Striker Posted October 7, 2014 Report Share Posted October 7, 2014 Cos = adjacent/hypotenuses So, Cos = a/2(3)^(1/2). That implies a(3)^(1/2)/6. If Cos = 2/3, then 2/3 = a(3)^(1/2)/6 implies 4 = a(3)^(1/2). That also implies a = 4/3^(1/2) or a = 4(3)^(1/2)/3. *sees above post* Oh come on now... Link to comment Share on other sites More sharing options...
hypebeast Posted October 7, 2014 Author Report Share Posted October 7, 2014 Okay guys I dont get it. Link to comment Share on other sites More sharing options...
Catman25 Posted October 7, 2014 Report Share Posted October 7, 2014 The angle doesn't matter in finding the ratios in this case, because they give you cos = 2/3, just using it for clarity's sake. So really 2/3=a/h so just solve for a, then use pythagorean to solve for the last side, o. Link to comment Share on other sites More sharing options...
hypebeast Posted October 7, 2014 Author Report Share Posted October 7, 2014 dude but 4sqrt(3)/3 ^ 2 is 5.33... Link to comment Share on other sites More sharing options...
Catman25 Posted October 7, 2014 Report Share Posted October 7, 2014 Yes it is, 5.33+20/3 = 12 = (2sqrt(3))^2 What part are you having trouble with? Link to comment Share on other sites More sharing options...
hypebeast Posted October 7, 2014 Author Report Share Posted October 7, 2014 Yes it is, 5.33+20/3 = 12 = (2sqrt(3))^2 What part are you having trouble with? The fact that in the test we arent allowed to use calculators, so results are usually exact. Like, thats an unusual result. Link to comment Share on other sites More sharing options...
Catman25 Posted October 7, 2014 Report Share Posted October 7, 2014 Well it's exact if you don't convert to decimals. For a radical, 4sqrt(3)/3 isn't bad at all. For what you had in the OP, that last "?" on the right side is just a 4, so it's not an overly ridiculous number. Link to comment Share on other sites More sharing options...
hypebeast Posted October 7, 2014 Author Report Share Posted October 7, 2014 Okay... You know what pisses me off? We have to leave it as it is, so for example if I were to do cosine myself it would be 4sqrt(3)/3 / 2sqrt(3). Idk if you get me, but since we dont use calculators we have a bunch of numbers on top of each other at all times. Link to comment Share on other sites More sharing options...
Catman25 Posted October 7, 2014 Report Share Posted October 7, 2014 Yeah I feel ya, improper fractions just take practice. What I usually do (you may or may not do this already) is take the first fraction and just multiply it by the reciprocal of the second, so 4sqrt(3)/3 * 1/2sqrt(3), and that should make it easier I think. Link to comment Share on other sites More sharing options...
goddamnit names are a pain Posted October 7, 2014 Report Share Posted October 7, 2014 Oh, I think I know this stuff... Hypotenuse= 2√3 Cosine= 2/3? or 2√3? Assuming the first... Adjacent/Hypotenuse=cosine Ok, solving for the adjacent is where I'm getting a bit lost, haha…a different scenario than I'm sort of used to atm, and it's late here... Pythagorean Theorem is then used….after this is calculated. So, our sides are= 2√3, _____, and ______. From there, we just use our known identities and such... Tangent= sin/cos Sin=Opposite/hypotenuse Cosine=given Cotangent= cos/sin Cosecant= hypotenuse/sin Secant= hypotenuse/cos Solve from there. Does this help? A calculator really isn't needed, and can throw off later calculations. Oh, I hope I didn't misread this, lol…and this is late... I messed up, reworking. Ehh…yeah, I shouldn't of really tried, lol... Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 Hey guys I got a surprise guess what it is yeah more trigonometry lol. This is a stupid one. Theres a right triangle with a perimeter of 36 cm. One of its sides (not the hypothenuse) is 12 cm. How do I determine the lenght of the other two sides? Link to comment Share on other sites More sharing options...
Catman25 Posted October 9, 2014 Report Share Posted October 9, 2014 You have two equations, you know: o^2+a^2=h^2 (opposite, adjacent, hypotenuse). The side length they give you can be either "o" or "a", doesn't matter, so let's use it for "o", thus 144+a^2=h^2. You also know the perimeter, so the side lengths should add up to 36: o+a+h=36, or 12+a+h=36. So the two equations are: 144+a^2=h^2 12+a+h=36 Solve for h or a, I solved for h. From the second equation we get h=24-a. Substitute this into the first eq. to get 144+a^2=(24-a)^2. This simplifies to 144+a^2=576-48a+a^2. a^2 cancels on both sides and we get 48a=432, so a=9. Plug into either equation to get h=15. 12+9+15=36, 144+81=225 Link to comment Share on other sites More sharing options...
Nathanael D. Striker Posted October 9, 2014 Report Share Posted October 9, 2014 Ignore me, since Catman keeps posting too fast. Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 You have two equations, you know: o^2+a^2=h^2 (opposite, adjacent, hypotenuse). The side length they give you can be either "o" or "a", doesn't matter, so let's use it for "o", thus 144+a^2=h^2. You also know the perimeter, so the side lengths should add up to 36: o+a+h=36, or 12+a+h=36. So the two equations are: 144+a^2=h^2 12+a+h=36 Solve for h or a, I solved for h. From the second equation we get h=24-a. Substitute this into the first eq. to get 144+a^2=(24-a)^2. This simplifies to 144+a^2=576-48a+a^2. a^2 cancels on both sides and we get 48a=432, so a=9. Plug into either equation to get h=15. 12+9+15=36, 144+81=225 Dude are you like Aristhotheles or something? Link to comment Share on other sites More sharing options...
Catman25 Posted October 9, 2014 Report Share Posted October 9, 2014 i used to enjoy math Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 i used to enjoy math Wait, why does this: 144+a^2=576-48a+a^2 Happen? How does it go from (24-a)^2 to 576-48a+a^2? Link to comment Share on other sites More sharing options...
Nathanael D. Striker Posted October 9, 2014 Report Share Posted October 9, 2014 i used to enjoy math I'm a math major. I would have gotten that, though I'm on my phone. XD @Hypebeast: (24-a)^2 = (24-a)(24-a) = 576-48a+a^2 Just some algebra there. Link to comment Share on other sites More sharing options...
Catman25 Posted October 9, 2014 Report Share Posted October 9, 2014 Wait, why does this: 144+a^2=576-48a+a^2 Happen? How does it go from (24-a)^2 to 576-48a+a^2? So it's a squared quantity, you need to expand it, you might've learned foil? So basically 24^2-24a-24a+a^2 @Striker: Math in college killed it for me lol Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 So it's a squared quantity, you need to expand it, you might've learned foil? So basically 24^2-24a-24a+a^2 Oh my god. Link to comment Share on other sites More sharing options...
Catman25 Posted October 9, 2014 Report Share Posted October 9, 2014 oh my god good, or oh my god bad Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 Oh my god as in realization. Link to comment Share on other sites More sharing options...
hypebeast Posted October 9, 2014 Author Report Share Posted October 9, 2014 Thanks to our lord and savior Catman25 I passed today's test. And actually I was one of the few people that knew how to solve it and I had to explain people how to do it lmao. Link to comment Share on other sites More sharing options...
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