I love that you include and explain all the rules that you use to solve problems. I've said that before but it's a major part of why I love your videos. Thank you!
@@satanssanta2478 Andy explained it in the video; an angle at the center of a circle has the same angle as the arc it contains, while an angle on the edge will have half that value
Yes the same thing struck me as well, the angle subtended by an arc at the centre is 2* the angle subtended by the same arc at the circumference of a circle.
I solved it a bit differently, by looking at the 30-60-90 triangle. Constructing a line L from the right angle to the midpoint of the hypotenuse creates an equilateral triangle, so L is equal to half the hypotenuse(also proved by the inscribed right triangle method cuz of radii). We know that the other side on the 50-50-80 triangle is also equal, so now we have an isosceles triangle in the middle. Then it’s only the matter of solving the corners and I got 40 degrees This method felt much more intuitive to me cuz I’m not very familiar with circle theorems but yours is genius!
Damn. I just typed a novel getting the to the solution the same way you did. I wish I saw yours first. Mine is much more confusing than this. Very well written.
Nice one, and a clever solution. As an engineer I thought of it as a surveyor and I went straight to trigonometry and a combination of the Pythagorean theorem and laws of sines and cosines to reach the same result.
The way to solve it with the circumscribed circle was pretty genius. I solved it by extending the two sides next to the side with two equal parts to make one big triangle with angles 50-60-70. From there, you can fiddle around with triangles and sides to get the desired angles, but it was a lot more work than your method.
That's a solid method. I actually agree that the inscribed angle concept is unnecessary here, the 30-60-90 triangle can be used to show that the right side of the quadrilateral is equal to both of the hashmarked congruent base segments. Then you have an isosceles triangle with a 60 degree peak, which is an equilateral triangle. That, plus recognizing the 50-50-80 isosceles triangle, gives us the two new "congruent radii" that he created with the inscribed angle bit. From there it's simple math on angles equaling 180.
Yiu can also solve using purely angles, sides, and triangle properties. The 50-80 triangle is isosceles, so it's righthand side is equal to the two marked halves of the base of the quadrilateral. As you demonstrated, the 60 angle creates a 30-60-90, and the right side of the quad is therefore one half of the base - so equal to the marked halves. If we run a line from the upper right vertex of the quad to the center of the base, it divides the 30-60-90 triangle into an equilateral (rt side) and isosceles (lt side). That new line will thus also be equal to the existing (almost) vertical line, creating a new isosceles triangle. With the 80 given, and the 60 from the equilateral triangle, that leaves 40 for the bottom angle of the new isosceles. The other two angles of the new isosceles must therefore each be 70. If the left hand angle (adjacent to the blue marked 50) is 70, and the red marked angle is 70, then our mystery green angle can only be 40. And just for fun, that original triangle with the green ? is ALSO an isosceles! 😊
I got to the same answer, but differently. Didn't know about the inscribed quadrilateral trick. In my solution, I drew a line completing the equilateral triangle. This makes a new isosceles triangle with the circle center and the two points at the top. The smallest angle can be known by subtracting the 60deg of the equilateral triangle from the known angle of 100, giving us 40. The other angles of this triangle are each (180 - 40)/2, or 70. This puts 70 degrees in the same triangle with the angle we're trying to solve. 180 - 70 - 70 gives us 40.
This took me forever to understand but I finally got it. I totally see what you did and I think that works but how do you know that the extra line you drew makes an equilateral triangle? I get the new line would be congruent or equal to radius but what about the 3rd side?
@@joegarcia4673 The short side of a 30-60-90 triangle is always half the hypotenuse, so once you solve those angles you know that side is equal to the tick marked lengths.
I think this can be done using just triangles. Once you have the 30° angle, the triangle it shares with the 60° angle gives you the 90° angle, and from that 30-60-90 triangle you can use that the short side is half the hypotenuse , i.e. the marked segment, let's call that length L. A 60° angled isosceles is an equilateral triangle, so we can draw a line from orange to green of length L. From the 50-50-80 isosceles triangle we also know that orange to blue is length L. That gives us an isosceles triangle (orange, blue, green) with a 40° (100-60) angle near orange, so the other two must be 70° (140°/2). Angle a + 30° (90° from the right angle - 60° from the equilateral) is 70°, so a is 40°.
I faules math in highschool, I found your videos while studying for a union carpenter test. I didn’t get in, but I passed the math test with flying colors. You made me love math
Woah man. I dodged the circles because I don’t know how to work with them. Love the logic you used though. I marked the same base angles. Next I used the 30 60 90 triangle properties to say the opposite leg on the right side equals x. With the split of the hypotenuse (2x) I found an equilateral triangle from the center to the top right angle. Because we proved there is an isosceles triangle on the left whose legs are equal in length to the equilateral on the right a new isosceles triangle is found with a 40° apex. 40° being the difference of 100° and the 60° equilateral angle found center bottom. With that new triangle we can find the missing 60° angle at the top. With that it’s a 180-70-60=50 for the missing angle. If that makes sense. You are right. It was fun.
I solved it using the following concepts 1.opposite angles 2. External angle of the triangle 3. The sum of all angles in a triangle is 180° 4. Cos(60) 5. Isosceles triangle 6. Equilateral triangle
I connected the middle where the 100° is, to the 90°, making a triangle out of the right side. Since the base is the radius and the newly made line is as well, we know that the triangle has 2 equal and therefore makes a 60 60 60 triangle. With the line i made in the beginning i made a triangle from center to x° to "?". Here i know the top right angle is 30° since 90-60, and that it is also equal sided. Therefore x°="?"+30. I use the fact that the top part of the triangle is 70°+"?"+x°=180. i the substituted for x° in and solve for "?". 70+30+?+?=180 2?=80 ?=40°
I didn't use arcs in my solution at all. I used the same geometric application that is used for finding the components of weight on a slope. I did this by making a vertical line up from the point with the 80 degree angle. I knew this would be 10 degrees left because the veritcal forms a right angle, I then connected a dotted horizontal line from the corner with the 50 degree mark to the vertical line to form another right triangle which had a known angle of 10 degrees between the given segment and the horizontal line. I knew that this horizontal line was cut by a transversal so it would be 80 - 10 = 70 degrees for the angle adjacent to the 50 degree angle. 180 - 70 - 70 = 40
Another way of solving it would be analyzing that the opposite angle of 70° must also be 70°, and since this opposite angle=70° then the other two angles must be 110°in order to complete the 360°. Since the 110° is supplementary to the other unknown angle inside the triangle, it should also be 70°. Then, 180°-70°-70°=40°, which is your answer! Good one!
I got most of the way, even noticed the isosceles triangle, but using a circle for the triangle never occurred to me, and i never even knew the rules for an inscribed quadrilateral. Super cool
every problem that has to fo with angles and triangles are always easier when you imagine a circle inscribed to that shaped. Took me 2 minutes to solve
i did it without all of those properties here is what i did please correct me if i wrong :) in the figure we can see we are given with one angle to be 60* and there is a triangle with angles 50* and 80* and the other angle in that triangle will be 50* (angle sum property) now we got one angle 60* and other one 50* and we have to find out x as we can see that by linear pair 180-70* = 110* so the value of angle(?)+angle(x) = 110 by exterior angle property and if subtract 110 from 180 then we can easily find the angle x and now as we know that angle x is 70* we can just add up the angles in the triangle which is 70*+70*+x = 180 x = 40 degrees :)))
man, i ended up going down the wrong path, i didnt see the circle, but saw the 30/60/90 triangle and the bisected hypotenuse, i was sure that it would be relevant that i could make my own line from the "circle center" to the 90 degree angle to make an equilateral triangle, but in the end it didnt get me anywhere.
Made it to the same point before the circle but didn't know what to do after. Even one length value would probably be enough to solve without the circle.
He didn't need to bring circles into this at all. You can figure out there's an isotopes triangle with no difficulty without knowing about right triangle hypotenouses being circle diameters. This is an overcomplication.
I just rotated my device and looked at the top right triangle and thought "hmm... Looks like an isosceles triangle" and then calculated the answer in my mind and got 40° 😂😂
The way i did it was finding out it could be inscribed in a circle, realising that the 2 points on the 50-50-80 triangle arc went to the circumference at ? and the centre at 80 degrees. Therefore ? was 1/2 of 80 degrees = 40 degrees (another circle theorem)
I didn’t see the semicircle. I created an equilateral triangle from the 90 degree angle to the 110 degree angle. From there I was able to gather the final triangle was an isosceles triangle with its two sides equal to the radius of the unobserved semicircle. My way seemed easier
I took the 30-60 right triangle proportions so the hypothenuse=2 and solved with laws of sines and cosines until I found that the last triangle was isocele and therefore has an angle of 40°
I got the answer in like 5 second cause I just looked that the small triangle has the equal long lines making the other corner (not ?) also be 70 and 70 + 70 is 140 and 180 - 140 is 40
I appreciate the struggle u did to solve it but, u don't need to struggle that much because, the triangle with ?, 70 degree angle is actually an isoceles triangle so the angle opposite to the 70 degree angle is also 70 degree and with some calculations: 70+70+? =180 140+? = 180 ? =180-140 ? = 40 degrees
Bro adding opposite side of a quadrilateral is 180 degree so 60° + 50° + x is equal to 180° so x is equal to 70° Now adding all the three angles of a triangle is 180 degrees so in the triangle where we have to find the angle we got two angles both are 70 degrees so so 70°+ 70°+ angle ' ? ' =180° So 140°- 180° = 40° and the question mark angle is 40 degree How exciting it is
indeed hard i do not new anything about the circles stuff and the oposite side of the quadrilateral but I hot all the other angles is there a solution using only the angles?
Maar je kan toch ook gewoon zien dat bij die hoek die rood is (70°) gelijk is aan die daar boven en de driehoek is altijd samen 180° dus dan doe je 180°-(70°X2) en dat is 40° dus zo kan je het toch ook oplossen?
Am I going crazy or does ?=10° also work? The triangle its in will make x=100° All the quadrilaterals will be satisfied All the triangles will be satisfied I'm blind to my error, please help!
I love that you include and explain all the rules that you use to solve problems.
I've said that before but it's a major part of why I love your videos. Thank you!
Yeah, it also helps the weak and the newbies.
Or, for the last one, you could use the fact that angle ? connects with the same arc as the angle of 80 degrees, making ? equal to 40 degrees.
This is such a cool solution!
@@satanssanta2478 Andy explained it in the video; an angle at the center of a circle has the same angle as the arc it contains, while an angle on the edge will have half that value
@@AndyMath please make a video about this alternative solution... I didn't get the explanation (I'm more visual than theorist)
Yes the same thing struck me as well, the angle subtended by an arc at the centre is 2* the angle subtended by the same arc at the circumference of a circle.
Yes please make a video explaining @cheesesentience alternative solution
How exciting this one is!
Every triangle can be inside a circle WOW! That is exciting
I solved it a bit differently, by looking at the 30-60-90 triangle. Constructing a line L from the right angle to the midpoint of the hypotenuse creates an equilateral triangle, so L is equal to half the hypotenuse(also proved by the inscribed right triangle method cuz of radii). We know that the other side on the 50-50-80 triangle is also equal, so now we have an isosceles triangle in the middle. Then it’s only the matter of solving the corners and I got 40 degrees
This method felt much more intuitive to me cuz I’m not very familiar with circle theorems but yours is genius!
Damn. I just typed a novel getting the to the solution the same way you did. I wish I saw yours first. Mine is much more confusing than this. Very well written.
I did the same thing! Was struggling to explain it in text, scrolled down to comments to see if anyone's described it already
Nice one, and a clever solution. As an engineer I thought of it as a surveyor and I went straight to trigonometry and a combination of the Pythagorean theorem and laws of sines and cosines to reach the same result.
The way to solve it with the circumscribed circle was pretty genius. I solved it by extending the two sides next to the side with two equal parts to make one big triangle with angles 50-60-70. From there, you can fiddle around with triangles and sides to get the desired angles, but it was a lot more work than your method.
That's a solid method. I actually agree that the inscribed angle concept is unnecessary here, the 30-60-90 triangle can be used to show that the right side of the quadrilateral is equal to both of the hashmarked congruent base segments. Then you have an isosceles triangle with a 60 degree peak, which is an equilateral triangle. That, plus recognizing the 50-50-80 isosceles triangle, gives us the two new "congruent radii" that he created with the inscribed angle bit. From there it's simple math on angles equaling 180.
Yiu can also solve using purely angles, sides, and triangle properties. The 50-80 triangle is isosceles, so it's righthand side is equal to the two marked halves of the base of the quadrilateral. As you demonstrated, the 60 angle creates a 30-60-90, and the right side of the quad is therefore one half of the base - so equal to the marked halves. If we run a line from the upper right vertex of the quad to the center of the base, it divides the 30-60-90 triangle into an equilateral (rt side) and isosceles (lt side). That new line will thus also be equal to the existing (almost) vertical line, creating a new isosceles triangle. With the 80 given, and the 60 from the equilateral triangle, that leaves 40 for the bottom angle of the new isosceles. The other two angles of the new isosceles must therefore each be 70. If the left hand angle (adjacent to the blue marked 50) is 70, and the red marked angle is 70, then our mystery green angle can only be 40. And just for fun, that original triangle with the green ? is ALSO an isosceles! 😊
That's the way I solve it too.
I got to the same answer, but differently. Didn't know about the inscribed quadrilateral trick.
In my solution, I drew a line completing the equilateral triangle. This makes a new isosceles triangle with the circle center and the two points at the top. The smallest angle can be known by subtracting the 60deg of the equilateral triangle from the known angle of 100, giving us 40. The other angles of this triangle are each (180 - 40)/2, or 70.
This puts 70 degrees in the same triangle with the angle we're trying to solve. 180 - 70 - 70 gives us 40.
This took me forever to understand but I finally got it. I totally see what you did and I think that works but how do you know that the extra line you drew makes an equilateral triangle? I get the new line would be congruent or equal to radius but what about the 3rd side?
This was my approach too!
@@joegarcia4673 The short side of a 30-60-90 triangle is always half the hypotenuse, so once you solve those angles you know that side is equal to the tick marked lengths.
@@bretonkyle ohhh I see thank youuu!!
hi andy. i like how you explain everything with so much detail.
I think this can be done using just triangles. Once you have the 30° angle, the triangle it shares with the 60° angle gives you the 90° angle, and from that 30-60-90 triangle you can use that the short side is half the hypotenuse , i.e. the marked segment, let's call that length L. A 60° angled isosceles is an equilateral triangle, so we can draw a line from orange to green of length L. From the 50-50-80 isosceles triangle we also know that orange to blue is length L. That gives us an isosceles triangle (orange, blue, green) with a 40° (100-60) angle near orange, so the other two must be 70° (140°/2). Angle a + 30° (90° from the right angle - 60° from the equilateral) is 70°, so a is 40°.
Or as someone else mentioned, once you know that x is 70°, angle a follows trivially from 180° - 70° - 70° = 40°
Yeah that's pretty not have to use the circle
Thank you Andy! Your videos have helped me a lot
I faules math in highschool, I found your videos while studying for a union carpenter test. I didn’t get in, but I passed the math test with flying colors. You made me love math
Hunnet. 😅
I heard it too he ain't slick 😂
absolutely amazing, your videos are always a highlight!
Alright, jokes aside, i truly find this solutions exciting.
Woah man. I dodged the circles because I don’t know how to work with them. Love the logic you used though.
I marked the same base angles.
Next I used the 30 60 90 triangle properties to say the opposite leg on the right side equals x. With the split of the hypotenuse (2x) I found an equilateral triangle from the center to the top right angle.
Because we proved there is an isosceles triangle on the left whose legs are equal in length to the equilateral on the right a new isosceles triangle is found with a 40° apex. 40° being the difference of 100° and the 60° equilateral angle found center bottom.
With that new triangle we can find the missing 60° angle at the top. With that it’s a 180-70-60=50 for the missing angle.
If that makes sense.
You are right. It was fun.
I solved it using the following concepts
1.opposite angles
2. External angle of the triangle
3. The sum of all angles in a triangle is 180°
4. Cos(60)
5. Isosceles triangle
6. Equilateral triangle
I connected the middle where the 100° is, to the 90°, making a triangle out of the right side. Since the base is the radius and the newly made line is as well, we know that the triangle has 2 equal and therefore makes a 60 60 60 triangle. With the line i made in the beginning i made a triangle from center to x° to "?". Here i know the top right angle is 30° since 90-60, and that it is also equal sided. Therefore x°="?"+30. I use the fact that the top part of the triangle is 70°+"?"+x°=180. i the substituted for x° in and solve for "?".
70+30+?+?=180
2?=80
?=40°
I didn't use arcs in my solution at all. I used the same geometric application that is used for finding the components of weight on a slope. I did this by making a vertical line up from the point with the 80 degree angle. I knew this would be 10 degrees left because the veritcal forms a right angle, I then connected a dotted horizontal line from the corner with the 50 degree mark to the vertical line to form another right triangle which had a known angle of 10 degrees between the given segment and the horizontal line. I knew that this horizontal line was cut by a transversal so it would be 80 - 10 = 70 degrees for the angle adjacent to the 50 degree angle. 180 - 70 - 70 = 40
Another way of solving it would be analyzing that the opposite angle of 70° must also be 70°, and since this opposite angle=70° then the other two angles must be 110°in order to complete the 360°. Since the 110° is supplementary to the other unknown angle inside the triangle, it should also be 70°.
Then, 180°-70°-70°=40°, which is your answer! Good one!
Didn't think I could solve this in less than 3 minutes just by imagining the shape in my mind.
i need to retake geometry i wouldve never thought of that!
Andy you Rock!
I got most of the way, even noticed the isosceles triangle, but using a circle for the triangle never occurred to me, and i never even knew the rules for an inscribed quadrilateral. Super cool
thanks for making math fun
This one was a really fun one. How a few simple ideas from geometry can lead to an easy solution.
I was proud of myself for brute forcing my way to the solution just to watch the video and feel like a barbarian. I was also off by 2 which also hurt.
Opposite angles of a quadrilateral have to add to 180. That would have made the last step easier.
Outstanding, love all your vodeos.
Wow! I guessed 40 right after just filling up all the angles using ASP, straight angle and VO angles 😂😂
every problem that has to fo with angles and triangles are always easier when you imagine a circle inscribed to that shaped. Took me 2 minutes to solve
I spent 10 minutes attempting to use trigonometric functions only to realize the solution was so simple lol 😂
0:07 160?
So close, figured the ?+90, i was close 😅
i did it without all of those properties
here is what i did please correct me if i wrong :)
in the figure we can see we are given with one angle to be 60*
and there is a triangle with angles 50* and 80* and the other angle in that triangle will be 50* (angle sum property)
now we got one angle 60* and other one 50* and we have to find out x as we can see that by linear pair 180-70* = 110* so the value of angle(?)+angle(x) = 110 by exterior angle property and if subtract 110 from 180 then we can easily find the angle x and now as we know that angle x is 70* we can just add up the angles in the triangle which is 70*+70*+x = 180
x = 40 degrees :)))
Clever boy
Clever and exciting! :) in a playful way!
I'd say it's interesting rather than exciting
Oh my God! I actually understood what you said in this one 😂
To me, this is easier than it looks
How exciting!
I got there using rules of similar triangles about the vertical angle, and the known ratio of the equivalent lengths from the isosceles triangle.
man, i ended up going down the wrong path, i didnt see the circle, but saw the 30/60/90 triangle and the bisected hypotenuse, i was sure that it would be relevant that i could make my own line from the "circle center" to the 90 degree angle to make an equilateral triangle, but in the end it didnt get me anywhere.
Made it to the same point before the circle but didn't know what to do after. Even one length value would probably be enough to solve without the circle.
Once you figure out that it’s a cyclic quadrilateral, it’s straightforward
He didn't need to bring circles into this at all. You can figure out there's an isotopes triangle with no difficulty without knowing about right triangle hypotenouses being circle diameters. This is an overcomplication.
And here's another thing, x = 70 (180 - 110 = 70), which means that the angle was in an isosceles triangle
I just rotated my device and looked at the top right triangle and thought "hmm... Looks like an isosceles triangle" and then calculated the answer in my mind and got 40° 😂😂
The way i did it was finding out it could be inscribed in a circle, realising that the 2 points on the 50-50-80 triangle arc went to the circumference at ? and the centre at 80 degrees. Therefore ? was 1/2 of 80 degrees = 40 degrees (another circle theorem)
You could also use external angle sum property
This one was FABOLOUS!!!
Wow!!! This one is indeed exciting!
1:20 “110 plus 💯 plus 60”
I didn’t see the semicircle.
I created an equilateral triangle from the 90 degree angle to the 110 degree angle. From there I was able to gather the final triangle was an isosceles triangle with its two sides equal to the radius of the unobserved semicircle.
My way seemed easier
Damn, I tried fitting it in an equilateral one, but no luck
I took the 30-60 right triangle proportions so the hypothenuse=2 and solved with laws of sines and cosines until I found that the last triangle was isocele and therefore has an angle of 40°
Because of the 70° angle***
1:50 "I wish I was high on POTenuse!" 😶🌫
I got the answer in like 5 second cause I just looked that the small triangle has the equal long lines making the other corner (not ?) also be 70 and 70 + 70 is 140 and 180 - 140 is 40
I appreciate the struggle u did to solve it but, u don't need to struggle that much because, the triangle with ?, 70 degree angle is actually an isoceles triangle so the angle opposite to the 70 degree angle is also 70 degree and with some calculations:
70+70+? =180
140+? = 180
? =180-140
? = 40 degrees
Bro adding opposite side of a quadrilateral is 180 degree so 60° + 50° + x is equal to 180° so x is equal to 70°
Now adding all the three angles of a triangle is 180 degrees so in the triangle where we have to find the angle we got two angles both are 70 degrees so so 70°+ 70°+ angle ' ? ' =180°
So 140°- 180° = 40°
and the question mark angle is 40 degree
How exciting it is
it was easily as the angle is in an isoscelens triangle
just
180-70-70=40
done
easy
How exciting !!!
1:51 self note: inscribed angle
I thought it was 60 tho since i assumed the bisector bisects its angle by half too, unless i remember maths wrong
Huh... I never would've thought of using a circumcised quadrilateral to figure out those angles. 😂😂😂
classic circumcised quadrilaterals
indeed hard
i do not new anything about the circles stuff and the oposite side of the quadrilateral
but I hot all the other angles
is there a solution using only the angles?
X=70, if anyone was wondering
We could have just used exterior and property in the small triangle
70 + ? = 110
? = 40
You just made it even complex 😢
Can you use cosines and sines laws, and assume that the || sides are 1 unit long?
Does a non-planar “triangle” on the surface of a sphere also have angles that add to 180o?
1st comment! How exciting!
I was this close
To solving it i got all except the last trianlge
Maar je kan toch ook gewoon zien dat bij die hoek die rood is (70°) gelijk is aan die daar boven en de driehoek is altijd samen 180° dus dan doe je 180°-(70°X2) en dat is 40° dus zo kan je het toch ook oplossen?
Bro I took a guess of 40 degrees before starting the video and it is correct 😂😂😂😂😂
Its simple angle chasing
Nice solution, could you do it without a circle?
Which software you use to make this type of solution?
i will be honest .....after finding all the angles except those 2 ...i make a guess of bw 40 or 50 ...i choose 40 ..and that was correct
Am I going crazy or does ?=10° also work?
The triangle its in will make x=100°
All the quadrilaterals will be satisfied
All the triangles will be satisfied
I'm blind to my error, please help!
why do you need to put it inside the/a circle?
Why you used Pythagoras theorem? You could be solved it whitout the theorem...
I got 30° ,tried it by different method😕
Easy 50
Comment
How hard can it be? Just use a protractor
🤔
I got everything up until inscribed shapes. FUCK