For extras on Q 16: • What Is The Most Compl... Geometry snacks books: www.tarquingro... Extras on Q24: • Most US College Studen... • Work rate problems
For question 22, I found a completely weird solution and it looks quite cool so I will put it below here: using the same chord same angle circle theorem rule, you can find a right angled isosceles triangle with the same chord where you find the side by the 1 1 sqrt2 triangle, getting the chord's length to be 2sqrt2 I then used sine rule to find another length of the original triangle and using area sine rule I found the answer.
For question 10 you can also write the angles as a ratio 40:72:108:140, which when simplified and summed gives 90 too (also thanks for all the videos, actually a life saver preparing for one of these!)
32:58 i have underlined -16 because i am a moron 😂 i love that ahahahah
For question 22, I found a completely weird solution and it looks quite cool so I will put it below here:
using the same chord same angle circle theorem rule, you can find a right angled isosceles triangle with the same chord where you find the side by the 1 1 sqrt2 triangle, getting the chord's length to be 2sqrt2
I then used sine rule to find another length of the original triangle and using area sine rule I found the answer.
I love how you explained 24
I was just about to comment that. Never seen anyone would define a task' lmao.🤣🤣🤣
Top Tip:
1/Total Time = 1/A + 1/B + 1/C ... 1/n
@@finnwilde oh yea that's a good one m8 thanks for sharing it
For question 10 you can also write the angles as a ratio 40:72:108:140, which when simplified and summed gives 90 too (also thanks for all the videos, actually a life saver preparing for one of these!)
u can also just find the HCF which is easily found to be 4, then divide 360 by 4 to get 90
Question 24 was brilliant
Q25, how do you know the red line coming from the length 1 line will meet the square's corner?
just a guess
He proves it after tho
This is amazing, thank you