I learned this a different way which was must easier for me so I'm going to share it here for those who are confused :) Sine Law: sin A/a = sin B/b = sin C/c (Where capital letters are the angles and lowercase letters are the corresponding side lengths) Ambiguous Case Of Sine Law • Happens when you're given two side lengths, but only one angle that is NOT in between the side lengths • Means there are two possible angles To solve an ambiguous case: 1. Find the unknown angle using sine law 2. Subtract the new angle from 180 degrees to get the second possible angle
No ... greater than or equal to is correct. Remember that SIDE a is opposite the angle. (I actually made a mistake in labelling the side in case 2 it's not side c but b)
hi ms, I was wondering, In mrkennedy’s pdf tb answer key solution for number 7, why did he do sin40 degrees = h/399.8? (I understand how 399.8 was found)
Well, if you got 399.8 that was the hard part! Now the problem comes down to finding the height of the building which he calls h and then you use sin40 = opp/hyp so that’s h/399.8 Good job!
Well done, thanks! On reflection we can't see you work out the solution for the two triangle case just before the end of the vid, your camera was positioned too high, the two triangle casd is not in view.
So sorry! It’s difficult to record from an iPhone! Hope it didn’t deter you from understanding the lesson. If I ever find some time in June just life I will redo it. 😊
I apologise, my previous question did not make sense. I really meant to ask you could you please show me where the hypotenuse is for triangle 1 & triangle 2 in the two triangle solution. I think I am on the right track but I just want to be sure. See the triangle I am working on here: drive.google.com/file/d/16QKqKuctq8wDyRwhGtkix1grkp8SdrP0/view?usp=drivesdk Thanks!
It is a bit difficult for sure. Maybe if you made yourself some triangles with toothpicks that would help visualize it better for you. And, perhaps watch it again?
I learned this a different way which was must easier for me so I'm going to share it here for those who are confused :)
Sine Law: sin A/a = sin B/b = sin C/c
(Where capital letters are the angles and lowercase letters are the corresponding side lengths)
Ambiguous Case Of Sine Law
• Happens when you're given two side lengths, but only one angle that is NOT in between the side lengths
• Means there are two possible angles
To solve an ambiguous case:
1. Find the unknown angle using sine law
2. Subtract the new angle from 180 degrees to get the second possible angle
Thank you! I learned that from her video too though, her diagram with the toothpicks really helped!
I learn more from these 20 minute videos than I do from my 2 hour classes at school lol
NO distractions on my videos (except the occasional ad) ... that helps.
Thank you for your videos! They're incredibly helpful, I am preparing for grade 11 next year!!
Nothing like advanced preparation! Good for you!! Don't forget to share the channel with your friends! : )
For in case 3 in 14:42, should it be "a" is smaller and equal to "b"? :)
No ... greater than or equal to is correct. Remember that SIDE a is opposite the angle. (I actually made a mistake in labelling the side in case 2 it's not side c but b)
hi ms,
I was wondering, In mrkennedy’s pdf tb answer key solution for number 7, why did he do sin40 degrees = h/399.8? (I understand how 399.8 was found)
Well, if you got 399.8 that was the hard part! Now the problem comes down to finding the height of the building which he calls h and then you use sin40 = opp/hyp so that’s h/399.8
Good job!
Ohhhh i see, thank you!
I keep getting sin(102) x 10/sin30 = 22.016 but in the video it says 19.5. Im so confused on what went wrong.
Hmmm did you figure it out? I’ve done it several ways and always get 19.5. Sin(30) is 1/2 so 10/(1/2) =20 so basically it’s just sin (102) x20
Well done, thanks! On reflection we can't see you work out the solution for the two triangle case just before the end of the vid, your camera was positioned too high, the two triangle casd is not in view.
So sorry! It’s difficult to record from an iPhone!
Hope it didn’t deter you from understanding the lesson. If I ever find some time in June just life I will redo it. 😊
@@mshavrotscanadianuniversit6234 Thanks!
I apologise, my previous question did not make sense. I really meant to ask you could you please show me where the hypotenuse is for triangle 1 & triangle 2 in the two triangle solution. I think I am on the right track but I just want to be sure. See the triangle I am working on here: drive.google.com/file/d/16QKqKuctq8wDyRwhGtkix1grkp8SdrP0/view?usp=drivesdk Thanks!
I’m sorry but I did not have access to this file. I tried on two devices. Also I have been frantically busy! 😳
@@mshavrotscanadianuniversit6234 No worries, thanks! I think I have worked it out!
2:24, I thought it was supposed to be sin30 in the denominator not sin3. ????
yes ... sin 30 degrees ... I actually say that and correct it if you watched the next few seconds : )
The last part got cut off
Sorry about that! That’s the problem when you don’t have a live audience! I’m sure your textbook will show you the final diagram 😊
@@mshavrotscanadianuniversit6234 Defenetiley. Thanks!
Can you please do question 7 & 8 page 319
I’ve done question 7 for you and will post it momentarily. See if that is enough to help you do 8 and let me know.
@@mshavrotscanadianuniversit6234 Thank you and I have a question why do I have to use Cos in question 10 to find the height?
mshavrot.pbworks.com/w/file/fetch/143661174/IMG_9605%202.pdf
Here is question 7 ... now try 8 again : )
@@mshavrotscanadianuniversit6234 Thank you so much
9:30
Very cryptic Josh ... I give up ... the stick rolled off the desk??
I didn't understand
It is a bit difficult for sure. Maybe if you made yourself some triangles with toothpicks that would help visualize it better for you. And, perhaps watch it again?