You made this more complicated than it needed to be. You could have used the fact that the complementary angle to the 60 degree angle is 120 which would have allowed you deduce most of the other angles without jumping through all the hoops that you did.
If we draw a segment from A perpendicular to BD and mark the point of intersection with the letter F. Then AF=a and BF=x. We also have that FD=2-x and FE=3-x. We set Tan45°=a/(3-x) and Tan60°=a/(2-x). Therefore a=(3+√3)/2 and x=(3-√3)/2. We mark the angle ABF as α. Finally Tanα=2+√3. Then α=tan-¹(2+√3). α=75°.
Line from B to line segment AD that is perpendicular to AD. 90, 60, 30 degree triangle. So 30 degrees is part of angle B. There's also a 90 degree angle on the other side and an isoceles triangle. So 90, 45, 45. 45 + 30 = 75 degrees.
2nd Method Geometry of Otrhocenter H, and circumscribed Quadraterals (we must bring the radius of circumscribed right quadraterals and using Bisector to Midle of Hypotenuse of a Right Triangle.Then Braking KHF angle in two pieces 90 -2x and 90-x and constract the equation giving angle x=15 deggres.Hence B=90 -15= 75 Deg).First we bring the three altitudes AF, BK and CN.They intersect to point H.
May i suggest two ways of soving this problem? But none of them is as easy and elegant as yours.FIRST using Trigonometry. Bring the altitude AF=h perpendicular to BC.Let FD=x. Then angle DAF=30 degrees. We use the COMMON h=AF two times as side of the two right triangles AFD and AFC, to express x.X=h*tan(30)= h*Sqrt(3)/3 (1) And x+1=EC=AE=h (2) from isosceles right triangle AFC,with acute angle C=45 deg. Hence h*sqrt(3)/3=h -1 (=x both) (3). Solvlning EQ(3) we have h - h*sqrt3/3 =1 ...h=(3+sqrt3)/2 (4). So x=h -1=....=(1+sqrt3)/2 And BF=2 -x=....=(3 - sqrt3)/2 (5). Now we'll calculate AB and then we'll apply LAW OF SINES to ABC triangle. AB^2=AF^2+BF^2 (PYTHAG)=((3+sqrt3)^2)/4+((3 -sqrt3)^2)/4=24/4=6 so AB=sqrt6 (6). LAW OF SINES >>> BC/sinA = AB/sinC so 3/sinA= sqrt6/((sqrt2)/2) so sqrt6*sinA= 3*sqrt2/2 so sinA=(3*sqrt2)/(2*sqrt2*sqrt3)=...=(sqrt3)/2.Hence
Interesting problem. It can be solved wo trigonometry , just using the easy theorem of Sum or 3 angles in every ( plane) triangle is equal yo 180°. Solvable by an 11 yo pupil ( in France), of course, a good pupil.
That only gets you so far. You can calculate the angle of 15° at the top of the right hand triangle without difficulty, sure, but then what? You need some other tools because in the left hand triangle, you still have two unknown angles. Two unknown angles doesn't help if you're *only* relying on 3 angles adding to 180° - there would be infinite possibilities. The extra piece of information is that the base of the triangle is 2 units which provides a unique case, so you need some trigonometry to link side lengths and angles, whether that be using cos 60° as in the video, or (as per other comments) using the sine rule. If you are 11 years old and looking at these kind of problems then well done, there's so much amazing maths ahead of you, so keep on learning!
I'll try to send you later a solution available by an 11 yo pupils. Imagine some of mine 12 yo found thé 24 isometries conserving the cube. Almost of this very good class solved the Theorem of the 2 Squares... But at the moment, on my mobile phone, it's uneasy to developp explainations.
@@tanelkagan well, though it's uneasy to explain geometry without drawing, I make an essay : let's call the [BC] the basis whose length is 2+1 = 3 ; Be A the opposite vertex and D € [BC],with BD = 2 Be I €[AD], ID =DC = 1.; then DIC is isocel in D and both angles DIC = DCI = 30° ( IDC = 180 - 60 , because B,D,C aligned) Be J , midpoint of [BD], then JID aequilateral and JID = 60° . As BD = 2 ID, then BI is the heigth of a triangle rectangle in I and IBC = 30° ( sum of angles). Thus, IBC is isocel in I and IB = IC . But, as IAC is also isocel in I (because AIC = 180 - 30 =150 and ACI = 45 -30 = 15°) then AIB is isocel in I too and IAB = IBA = (180 - 90) : 2 = 45°, then BAC = 30 + 15 = 45° and ABC =45 + 30 = 75° thus, only using sum of angles = 180° and easy properties about isocel and aequilateral allow t solve this nteresting problem, just calculus...without trigonometry. Sorry for the notations , but you can draw from my text and it works...
You made this more complicated than it needed to be. You could have used the fact that the complementary angle to the 60 degree angle is 120 which would have allowed you deduce most of the other angles without jumping through all the hoops that you did.
Think you meant "supplementary"
Shall thank you to suggest the easier method, at least the basic steps of the proof
ADC=120°, DAC=180°-120°-45°=15°...
If we draw a segment from A perpendicular to BD and mark the point of intersection with the letter F. Then AF=a and BF=x. We also have that FD=2-x and FE=3-x. We set Tan45°=a/(3-x) and Tan60°=a/(2-x). Therefore a=(3+√3)/2 and x=(3-√3)/2. We mark the angle ABF as α. Finally Tanα=2+√3. Then α=tan-¹(2+√3). α=75°.
I ended up using arctangent too. And I kept thinking to myself "there's gotta be an easier way!" LOL
Yes that is my solution also
Line from B to line segment AD that is perpendicular to AD. 90, 60, 30 degree triangle. So 30 degrees is part of angle B. There's also a 90 degree angle on the other side and an isoceles triangle. So 90, 45, 45. 45 + 30 = 75 degrees.
Drop perpendicular and play with tangents
tan(45) = y/(3-x)
tan(60) = y/(2-x)
tan(theta) = y/x
شكرا لكم على المجهودات
يمكن استعمال المسقط العمودي لDعلى(AC)
2nd Method Geometry of Otrhocenter H, and circumscribed Quadraterals (we must bring the radius of circumscribed right quadraterals and using Bisector to Midle of Hypotenuse of a Right Triangle.Then Braking KHF angle in two pieces 90 -2x and 90-x and constract the equation giving angle x=15 deggres.Hence B=90 -15= 75 Deg).First we bring the three altitudes AF, BK and CN.They intersect to point H.
It can be easily done using m-n theorem
May i suggest two ways of soving this problem? But none of them is as easy and elegant as yours.FIRST using Trigonometry. Bring the altitude AF=h perpendicular to BC.Let FD=x. Then angle DAF=30 degrees. We use the COMMON h=AF two times as side of the two right triangles AFD and AFC, to express x.X=h*tan(30)= h*Sqrt(3)/3 (1) And x+1=EC=AE=h (2) from isosceles right triangle AFC,with acute angle C=45 deg. Hence h*sqrt(3)/3=h -1 (=x both) (3). Solvlning EQ(3) we have h - h*sqrt3/3 =1 ...h=(3+sqrt3)/2 (4). So x=h -1=....=(1+sqrt3)/2 And BF=2 -x=....=(3 - sqrt3)/2 (5). Now we'll calculate AB and then we'll apply LAW OF SINES to ABC triangle.
AB^2=AF^2+BF^2 (PYTHAG)=((3+sqrt3)^2)/4+((3 -sqrt3)^2)/4=24/4=6 so AB=sqrt6 (6).
LAW OF SINES >>> BC/sinA = AB/sinC so 3/sinA= sqrt6/((sqrt2)/2) so
sqrt6*sinA= 3*sqrt2/2 so sinA=(3*sqrt2)/(2*sqrt2*sqrt3)=...=(sqrt3)/2.Hence
LOVE YOUR VIDEOS
Excellent
Interesting problem. It can be solved wo trigonometry , just using the easy theorem of Sum or 3 angles in every ( plane) triangle is equal yo 180°. Solvable by an 11 yo pupil ( in France), of course, a good pupil.
That only gets you so far. You can calculate the angle of 15° at the top of the right hand triangle without difficulty, sure, but then what? You need some other tools because in the left hand triangle, you still have two unknown angles. Two unknown angles doesn't help if you're *only* relying on 3 angles adding to 180° - there would be infinite possibilities.
The extra piece of information is that the base of the triangle is 2 units which provides a unique case, so you need some trigonometry to link side lengths and angles, whether that be using cos 60° as in the video, or (as per other comments) using the sine rule.
If you are 11 years old and looking at these kind of problems then well done, there's so much amazing maths ahead of you, so keep on learning!
I'll try to send you later a solution available by an 11 yo pupils. Imagine some of mine 12 yo found thé 24 isometries conserving the cube. Almost of this very good class solved the Theorem of the 2 Squares... But at the moment, on my mobile phone, it's uneasy to developp explainations.
@@tanelkagan well, though it's uneasy to explain geometry without drawing, I make an essay :
let's call the [BC] the basis whose length is 2+1 = 3 ; Be A the opposite vertex and D € [BC],with BD = 2 Be I €[AD], ID =DC = 1.; then DIC is isocel in D and both angles DIC = DCI = 30° ( IDC = 180 - 60 , because B,D,C aligned)
Be J , midpoint of [BD], then JID aequilateral and JID = 60° . As BD = 2 ID, then BI is the heigth of a triangle rectangle in I and IBC = 30° ( sum of angles). Thus, IBC is isocel in I and IB = IC . But, as IAC is also isocel in I (because AIC = 180 - 30 =150 and ACI = 45 -30 = 15°) then AIB is isocel in I too and IAB = IBA = (180 - 90) : 2 = 45°, then BAC = 30 + 15 = 45° and ABC =45 + 30 = 75°
thus, only using sum of angles = 180° and easy properties about isocel and aequilateral allow t solve this nteresting problem, just calculus...without trigonometry. Sorry for the notations , but you can draw from my text and it works...
75 degree. mindculus.
More than likely it is 75 since 60 is 15 more than 45: 45, 60, 75.
Спасибо.
75° Спасибо)
180 - 60 - 45 = 75
75°
This 75
Вашу задачу своровал русский боогер.