Seriously though dude...i'm 40 years of age and have never understood the jumbled signs and symbols i knew as math but due in large part to your vids and some other books i'm studying i'm really getting a handle on material that mystified me in the past...i'm still waiting for that "Eureka" moment that happened to you in the library but i'm a keep on keeping on....
They're functions that relate the sides of a triangle (right triangle) to it's angles. It's a ratio really. There is a lot more to it and it will take a whole series to cover it in detail, which i do plan on doing, but again, it will be quite some time before i get to it... sorry :(
I keep trying and struggling with one of your exercise problems on website and cannot figure out why answer is not same as yours. It refers to solving using the Pythagorean theorem on problem 2a. where a=5, b=5 and c or x is the unknown. Your answer was 5 time square root of 2. I came up with (on calculator) 7.071. The very next problem 2b. comes up with a similar result of 19.21. Why is 7.071 incorrect?
7.071 is not incorrect. It's the correct answer. 7.071... is equal to 5 times square root of 2. The only difference is that 5 times the square root of 2 is an exact answer while 7.071... is approximate since the square root of 2 is an irrational number (see series II for solving radicals).
->when in a right angle triange there is a 30 degree angle then the opposite side of this angle is equal to the half of the hypotenuse of this triangle.
Wait an minute...so now if i wanted i could use the 30 degree angle to solve x by using Cos once again...7 being hyp and x being adj?
MAJESTIcag7 4 months ago
Yup :)
chychochycho 4 months ago
@chychochycho Now i'm getting somewhere : )
Seriously though dude...i'm 40 years of age and have never understood the jumbled signs and symbols i knew as math but due in large part to your vids and some other books i'm studying i'm really getting a handle on material that mystified me in the past...i'm still waiting for that "Eureka" moment that happened to you in the library but i'm a keep on keeping on....
PEACE
MAJESTIC
MAJESTIcag7 4 months ago
Comment removed
MAJESTIcag7 4 months ago
Ok so...if i were to use the 60 degree angle to solve for x i would use Sin?
MAJESTIC
MAJESTIcag7 4 months ago
yup :)
chychochycho 4 months ago
@chychochycho Nice!!!And in the same vein...if i were to chose the 30 degree angle to solve for y i would use the.....Tan?
MAJESTIcag7 4 months ago
you could use sin with 7, or, now that you have x, you could use tan with x.
chychochycho 4 months ago
@chychochycho Ok...i got ya...i guess with enough practice which function to use'll become 2nd nature.
MAJESTIcag7 4 months ago
It will :)
chychochycho 4 months ago
@chychochycho So if i'm understanding it right you can use the sides as well as the angles with the fuctions of Sin,Cosine and Tangent?
MAJESTIcag7 4 months ago
Remember, you can only take the sin, cos, and tan of an angle.
chychochycho 4 months ago
you really needed the calculator to know that cos 60 is 1/2? come on. You are teaching maths and don't know your trig?
mikeyz75 9 months ago
I just wanted to know how to do, "x^2+(x+2)^2=100".
Austin101123 1 year ago
@Austin101123 the answers are 6 and -8, just use foil and the quadratic formula, but it probably doesn't matter anymore :p
Drummaboyy13 8 months ago
@Drummaboyy13 true i found out how to do that the next day in class lol
Austin101123 8 months ago
epic got a100 on my last test in college . THESE LESSON ARE EPIC XXXXXXXXX10
holy3 1 year ago
I like EPIC :)
and congrads on the 100. Job well done :)
chychochycho 1 year ago
i dont get how you get the 0.5, i wanna know how to do it without the calculator
mixolidianmode 1 year ago
i dont get what the sino, coso and tano is.. is it like a formula?
mixolidianmode 1 year ago
They're functions that relate the sides of a triangle (right triangle) to it's angles. It's a ratio really. There is a lot more to it and it will take a whole series to cover it in detail, which i do plan on doing, but again, it will be quite some time before i get to it... sorry :(
chychochycho 1 year ago
you should do word problems of this also after you are done with the basics like you said..
mixolidianmode 1 year ago
I keep trying and struggling with one of your exercise problems on website and cannot figure out why answer is not same as yours. It refers to solving using the Pythagorean theorem on problem 2a. where a=5, b=5 and c or x is the unknown. Your answer was 5 time square root of 2. I came up with (on calculator) 7.071. The very next problem 2b. comes up with a similar result of 19.21. Why is 7.071 incorrect?
carvingmadness 2 years ago
7.071 is not incorrect. It's the correct answer. 7.071... is equal to 5 times square root of 2. The only difference is that 5 times the square root of 2 is an exact answer while 7.071... is approximate since the square root of 2 is an irrational number (see series II for solving radicals).
chychochycho 2 years ago
This comment has received too many negative votes show
look like a terrorist
Ezekiel160 2 years ago
@Ezekiel160 WTF lol, that was so fuckin rude he's a math magician, terrorist doesn't even have internet so I'd suggest you shut up.
mqg96 2 years ago
Thank you!!!!!
thefragile5qh5 3 years ago
you're welcome :)
chychochycho 3 years ago
another solution to this problem.
THEOREM, (proved in the elements of euklid)
->when in a right angle triange there is a 30 degree angle then the opposite side of this angle is equal to the half of the hypotenuse of this triangle.
As far as your example......
ABC->right angle triangle
B=30, so x=hyp/2 x=7/2 x=3,5 (where,x=b)
pozerasssss 3 years ago
my pleasure :)
chychochycho 4 years ago