couldn’t you have … couldn’t you have use the law of cosines on the first problem?
lets call the ship point c
(CB)^2 = 5^2 + 10^2 – 2(5)(10)cos15degree
any videos of a … any videos of a right-angled triangle being cut into half(but not seperated): having only values for only 1 of 2
seperated triangles.
e.g:
tri ABD, where ABD = 90degrees
A produces a line where it meets at BD, calling it C.
The line AC, hence, made 2 triangles in 1.
Having AB=12, BC=5, AC=13, find cosACD.
hope that explains the question, any help? i’m stuck.
Lolz I almost got … Lolz I almost got lost on the y-part cuz I thought that the equal(=) sign looked like a minus(-) @_@
By the way Khan how come you like skipping steps and just subbing it in the end?
if you don’t mind … if you don’t mind me saying something…
it would be less confusing for people watching this if you finished the equation of x and y and z first before you continue. So people could get a better understanding of how your completing these equations.
Think of it like … Think of it like this. What is (x+5)^2?
Because (x+5)^2 is the same thing as:
(x+5) * (x+5)
Therefore expanding it gives you:
x^2+10x+25
An easy trick is to square the first term, then multiply the terms and double it, then square the last term.
So
(10-5cos15)
Expanded= 100-100cos15-25cos(15)^2
If it’s still tough, let (cos15) = x and do it like so:
(10-5x)^2
=100-100x-25x^2
=100-100(cos15)-25(cos15)^2
I got confused at … I got confused at the part where you said “if that confuses you” how do you get 100 – 100 cos I understand 5 sin 15 squared is = 25 sin 15 them + (10 squared) 100 then confused at – 100 cos 15?
This problem should … This problem should be easy to solve using Law of Cosines because you are given the answer to one angle and two legs; therefore, you can use a^2 = b^2+c^2-2bc*CosA. If A =15, b=5km and c=10km then take the square root of 5^2+10^2-2(5)(10)*Cos(15) to get about 5.3 km.
Thus the boat needs to travel about 5.3 km after changing heading by about 14 degrees to get back on course. The 14 degrees may be found using Law of Sines; a/SinA=b/SinB=c/SinC.
The first part of a problem when the captain of a ship goes off track.
March 6th, 2010 at 12:28 pm
I LOVE YOU!!!! …
I LOVE YOU!!!! Thank you soo much! your work is amazing i really appreciate this. Thank You
March 6th, 2010 at 12:28 pm
couldn’t you have …
couldn’t you have use the law of cosines on the first problem?
lets call the ship point c
(CB)^2 = 5^2 + 10^2 – 2(5)(10)cos15degree
March 6th, 2010 at 12:28 pm
thank you SO much …
thank you SO much for this. saved my butt.
March 6th, 2010 at 12:28 pm
I dont understand …
I dont understand what he does from 8:05 to 9:45 … he squares the yellow writing i get that but what did he do with the red?
March 6th, 2010 at 12:28 pm
Ahh i see, Im quite …
Ahh i see, Im quite happpy i noticed that though
March 6th, 2010 at 12:28 pm
@Catkill666 …
@Catkill666 Technically yes, he could have, but I don’t think he taught the Law of Cosines yet when he made the video.
March 6th, 2010 at 12:28 pm
I would hate to sit …
I would hate to sit in a classroom with this person for more than 30 minutes
March 6th, 2010 at 12:28 pm
Couldnt you just …
Couldnt you just use the cosine rule?
March 6th, 2010 at 12:28 pm
You are my hero.
You are my hero.
March 6th, 2010 at 12:28 pm
any videos of a …
any videos of a right-angled triangle being cut into half(but not seperated): having only values for only 1 of 2
seperated triangles.
e.g:
tri ABD, where ABD = 90degrees
A produces a line where it meets at BD, calling it C.
The line AC, hence, made 2 triangles in 1.
Having AB=12, BC=5, AC=13, find cosACD.
hope that explains the question, any help? i’m stuck.
March 6th, 2010 at 12:28 pm
Got lost on that …
Got lost on that one! Will watch again and again and again
March 6th, 2010 at 12:28 pm
Thank you so much, …
Thank you so much, this video was really helpful, you explained it alot more simplier than my math teacher -_-
March 6th, 2010 at 12:28 pm
Lolz I almost got …
Lolz I almost got lost on the y-part cuz I thought that the equal(=) sign looked like a minus(-) @_@
By the way Khan how come you like skipping steps and just subbing it in the end?
March 6th, 2010 at 12:28 pm
theres an h in …
theres an h in hypotenuse
March 6th, 2010 at 12:28 pm
A new side of a …
A new side of a triangle: The Oppotenuse!
Thank you, Sal.
March 6th, 2010 at 12:28 pm
if you don’t mind …
if you don’t mind me saying something…
it would be less confusing for people watching this if you finished the equation of x and y and z first before you continue. So people could get a better understanding of how your completing these equations.
March 6th, 2010 at 12:28 pm
Think of it like …
Think of it like this. What is (x+5)^2?
Because (x+5)^2 is the same thing as:
(x+5) * (x+5)
Therefore expanding it gives you:
x^2+10x+25
An easy trick is to square the first term, then multiply the terms and double it, then square the last term.
So
(10-5cos15)
Expanded= 100-100cos15-25cos(15)^2
If it’s still tough, let (cos15) = x and do it like so:
(10-5x)^2
=100-100x-25x^2
=100-100(cos15)-25(cos15)^2
March 6th, 2010 at 12:28 pm
I got confused at …
I got confused at the part where you said “if that confuses you” how do you get 100 – 100 cos I understand 5 sin 15 squared is = 25 sin 15 them + (10 squared) 100 then confused at – 100 cos 15?
March 6th, 2010 at 12:28 pm
this is good, …
this is good, theres some other good videos on tutormetvdotcom. It helped me with my midterms
March 6th, 2010 at 12:28 pm
I think it is does …
I think it is does make sense and it is certainly acceptable given the context; namely, the Law of Cosines and Law of Sines.
Is it more correct to say “you” than “u”? Both are correct as long as your communication is understood by its receiver.
Otherwise, I am missing the reason why it is does not make sense.
March 6th, 2010 at 12:28 pm
u said A=15, b=5km …
u said A=15, b=5km and c=10km this is does not make sense.
instead say angle A = 15 deg, AC = 5 km, and AB = 10km
March 6th, 2010 at 12:28 pm
without knowin this …
without knowin this is a trig course though how would you think of usin trig
March 6th, 2010 at 12:28 pm
This problem should …
This problem should be easy to solve using Law of Cosines because you are given the answer to one angle and two legs; therefore, you can use a^2 = b^2+c^2-2bc*CosA. If A =15, b=5km and c=10km then take the square root of 5^2+10^2-2(5)(10)*Cos(15) to get about 5.3 km.
Thus the boat needs to travel about 5.3 km after changing heading by about 14 degrees to get back on course. The 14 degrees may be found using Law of Sines; a/SinA=b/SinB=c/SinC.
March 6th, 2010 at 12:28 pm
you lost me at the …
you lost me at the drawing the line down the middle of triangle part
March 6th, 2010 at 12:28 pm
that is where the …
that is where the cosine rule came from………ㅡㅡ;;