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4:30 I did smush the two triangles together, continuing the vertex of length 3 up until a 90deg corner would cross over to the junction of the vertexes of length 12 and 13. From there I had two similar triangles and solved for the total length, which was the same as my opposite length for the angle A+B, then divided by 13, the hypotenuse, and got the same answer. How exciting.
I was gonna say, a great way to solve this is to scale the triangles into a Heronian triangle. Since you're pairing the small angle of the 3 4 5 with the large acute angle of the 5 12 13 you scale the 3 4 5 up by 4 to get 12 16 20. Then you glue the 12s together. 12 is the height of the new triangle, the base is 5 + 16 = 21, and the other sides are 12 and 20. Roll it over so now the base is 20. By using area = bh/2, we can find the height off the base 20 is 63/5. So let;s scale it by 5. The base is now 100, the other sides are 65 and 105. The height is 63. And we see the sine of the original angle, opposite the 105 side, has to be 63/65.
Great minds think alike. I did it exactly the same. I even tried to squish the two right-triangles together, before using the formula for the sine of a sum of angles.
Who else hates sin^-1? It looks like we should be raising it to the power -1. I prefer the arc sine, or asin, because then if you do exponentiate a trig function it's clearer.
I solved it graphically by starting with joining the two triangles together. Then I extended the segment of length 12 and the segment of length 4 until they intersected, creating another triangle with the same proportions as the 3-4-5 triangle, only that one being 2.25-3-3.75. Extend the line segment of length 4 the other direction and add a perpendicular line to make a big right triangle with the same 3-4-5 proportions, only this time being 9.45-12.6-15.75. Sin of the angle is then 12.6/13.
another method is with coordinate geometry. you have 2 right triangles like you showed with sides (3,4,5) and (5,12,13). Start with the (3,4,5) and place it starting at the origin with vertices at (0,0), (4,0), and (4,3) now imagine a point at (x,y) which forms the (5,12,13) triangle using the side from (0,0) to (4,3). To find x,y we can solve x^2+y^2=13^2 (x-4)^2+(y-3)^2=12^2 giving us x=-16/5 and y=63/5 Now we also know that y=13*sin(theta) where theta=atan(3/4)+acos(5/13) thus sin(theta)=y/13=63/65
I took a slightly more roundabout route, converted everything inside the [ ] to sin⁻¹ terms, then used the identity sin⁻¹a + sin⁻¹b = sin⁻¹{a√(1 - b²) + b√(1 - a²)}. Nice that the answer is also a Pythagorean triple (16, 63, 65).
@@ValKS-0 A couple of weeks ago, when I came across the phrase (in a restoration video, of all things), and didn't know what it was, I immediately searched for it on Google, and immediately found out what it was. Copy > paste > click > 'oh, OK'.
@@Grizzly01-vr4pn I recall hearing that it was customary to use Sin^-1 with triangles and degrees, and arcsin with circles and radians. It was never a big deal not to follow that, though.
Salt On Ham Can Always Help The Ordinary Appetite.
Like an army, a mathematician marches on their stomach! 🤣🤣
Some old hippie caught another hippee tripping on acid
When mnemonic is harder to memorise than just memorising it
Some old hippie came around here tripping on acid
That was a brilliant amount of trig.
I love this channel!
"when am i ever going to use this" duh. when you find a math RUclipsr and decided to binge all his content. obviously.
Loving the trig videos. I didn't learn anything more advanced than algebra in school, so it's fun to see what I missed out on.
4:30 I did smush the two triangles together, continuing the vertex of length 3 up until a 90deg corner would cross over to the junction of the vertexes of length 12 and 13. From there I had two similar triangles and solved for the total length, which was the same as my opposite length for the angle A+B, then divided by 13, the hypotenuse, and got the same answer. How exciting.
I was gonna say, a great way to solve this is to scale the triangles into a Heronian triangle. Since you're pairing the small angle of the 3 4 5 with the large acute angle of the 5 12 13 you scale the 3 4 5 up by 4 to get 12 16 20. Then you glue the 12s together. 12 is the height of the new triangle, the base is 5 + 16 = 21, and the other sides are 12 and 20. Roll it over so now the base is 20. By using area = bh/2, we can find the height off the base 20 is 63/5. So let;s scale it by 5. The base is now 100, the other sides are 65 and 105. The height is 63. And we see the sine of the original angle, opposite the 105 side, has to be 63/65.
Great minds think alike. I did it exactly the same. I even tried to squish the two right-triangles together, before using the formula for the sine of a sum of angles.
Who else hates sin^-1? It looks like we should be raising it to the power -1. I prefer the arc sine, or asin, because then if you do exponentiate a trig function it's clearer.
I always specify arcsin or csc because of this. Got me funny looks from my classmates
Same. I always prefer using cosec
And better yet, Sohcahtoa helped Lewis & Clark get through The Northwest Passage!
Which movie
@@BadakMahashay "Weekend At Bernie's".
@@joeschmo622 thx
That brought back some memories, not all of which were as much fun as watching your solution. Tthanks Andy!
I solved it graphically by starting with joining the two triangles together. Then I extended the segment of length 12 and the segment of length 4 until they intersected, creating another triangle with the same proportions as the 3-4-5 triangle, only that one being 2.25-3-3.75. Extend the line segment of length 4 the other direction and add a perpendicular line to make a big right triangle with the same 3-4-5 proportions, only this time being 9.45-12.6-15.75. Sin of the angle is then 12.6/13.
another method is with coordinate geometry.
you have 2 right triangles like you showed with sides (3,4,5) and (5,12,13). Start with the (3,4,5) and place it starting at the origin with vertices at (0,0), (4,0), and (4,3)
now imagine a point at (x,y) which forms the (5,12,13) triangle using the side from (0,0) to (4,3). To find x,y we can solve
x^2+y^2=13^2
(x-4)^2+(y-3)^2=12^2
giving us x=-16/5 and y=63/5
Now we also know that y=13*sin(theta) where theta=atan(3/4)+acos(5/13)
thus sin(theta)=y/13=63/65
Some Old Hen Caught Another Hen Taking Oats Away
Love how Vin's channel is growing so he can keep giving us this dope content
Perfect! Right before i take the CSET
This is what I do in 12th class Indian maths
Made me look! And the thumbnail is right, it was fun.
One of the best questions 😮❤❤
U make me fall in love with maths
SOH-CAH-TOA, yuck ! I like
Sin Cos Tan
“Oliver And Olivia
Have Hairy Ankles Sin Cos Tan”
I took a slightly more roundabout route, converted everything inside the [ ] to sin⁻¹ terms, then used the identity sin⁻¹a + sin⁻¹b = sin⁻¹{a√(1 - b²) + b√(1 - a²)}.
Nice that the answer is also a Pythagorean triple (16, 63, 65).
Can you do topological problems
Is there any place online where you can learn and practice the basics of trigonometry? I am having trouble understanding this stuff
How Exciting! 😁
loved it!
Great stuff. Love it. Do you like doing proofs?
kind of a bummer that there is no pure geometric way to solve this one
Perfect ❤
Most of your videos are of solving 10th,11th and 12th class ncert problems of indian students.😊😊
Only 6/65 away from a nice fraction
Redundant title, since trig is inherently fun 🤓
Doubly redundant as its also inherently challenging 😢
Cos Sin Tan Tan
How you made this video?
hello Andy from cowboy bepop
Holy shit i got it right
Wow
I memorized ''Hawk Tuah'' very quickly, but I can't remember ''SOH-CAH-TOA''. Can anyone explain to me, why it is like this?
What is hawk tuah?
@@ValKS-0 A couple of weeks ago, when I came across the phrase (in a restoration video, of all things), and didn't know what it was, I immediately searched for it on Google, and immediately found out what it was.
Copy > paste > click > 'oh, OK'.
I have never seen sin^(-1) being equal to arcsin.. Looks weird and confusing
It's been around for years. It's how we were taught to write inverse trig functions at school (UK, ⁓ 40 years ago).
@@Grizzly01-vr4pn I recall hearing that it was customary to use Sin^-1 with triangles and degrees, and arcsin with circles and radians. It was never a big deal not to follow that, though.
No 69? Not nice.
EARLY!