Can you find X+Y=? | Two cool methods | (Trigonometry and Geometry) |

[@ybodoN]

The first method is so beautiful that I had to watch it twice 🤩

[@PreMath]

Super! ❤️🌹

[@sail2byzantium]

It was very interesting to see--I confess I had to watch it twice. I wish I could be that clever. I was very literalist in doing the Pythagorean theorem followed by some trig work.

[@sail2byzantium]

Did the trig version--doing the tangents of both angles: x = 4/3 = 53.1301 . . . degrees and y = 7/1 = 81.8698 . . . degrees. 53.1301 . . . degrees + 81.8698 . . . degrees = 135 degrees.

[@anthonycheng1765]

tan(x)=4/3,tan(y)=7,tan(x+y)=(tan(x)+tan(y))/(1-tan(x).tan(y))=(4/3+7)/(1-4/3*7)=-1,x+y=tan(-1)=3*pi/4

[@johnbrennan3372]

Clear and concise

[@PreMath]

Thanks 🌹

[@TheDHemant]

I also thought of this method and got the answer but this is 11 standard's method. The premature is trying to explain it based on school maths. Undoubtedly he is great!!!

[@Irishfan]

The simpler solution would be to find the arctangent of both angles and angles them together. Arctangent 4/3 = 53.13° and Arctangent 7/1 = 81.87° then ad the angles 53.13° + 81.87° = 135° much much simpler!

[@montynorth3009]

Another trig.method using tangents to find the angles X & Y then adding them together. Tan X = 4/3 = 1.333. (Tan -1) 1.333 = 53.1301. Tan Y = 7. (Tan-1) = 81.8699. X + Y = 53.1301 + 81.8699 = 135.

[@PreMath]

Thanks for sharing! ❤️

[@juanalfaro7522]

I would have used the tangent of (x+y) in the 2nd method; it's much faster. The result is (-1) for which you infer the angle is 135 degrees (not 315 or -45 degrees), since the sum of 2 angles in the 1st quadrant is an angle not past the 2nd quadrant.

[@PreMath]

Great! Thanks ❤️

[@jimlocke9320]

Solution by tangent sum of angles formula: tan(x + y) = (tan(x) + tan(y))/(1 - tan(x)tan(y)). Here, tan(x) = 4/3, tan(y) = 7/1 = 7. tan(x + y) = (4/3 + 7)/(1 - (4/3)(7)) = (25/3)/(1 - 28/3) = (25/3)/(-25/3) = -1. arctan(-1) = 135°, so x + y = 135°.

[@PreMath]

Thanks ❤️

[@tombufford136]

At a quick glance, The Red outlined right angled triangle has sides 3,4 and 5 (Pythagorean triple). Then x = sin ^-1(4/5) = 53.13 and y = tan^-1(7/1) =81.87 then x + y = 53.13 + 81.87 = 135. Looking forward to the Video.

[@tellerhwang364]

Geometry puzzle method 1.use 2△ABC and 1△BDE △CDK,△ABC congruence △KHA,△BDE congruence →BDKH rectangle 2.CK=AC,ACK=90 △ACK lsosceles right triangle →CAK=45,CAB=x,KAH=y →x+y=180-45=135😊

[@PreMath]

Thanks ❤️

[@kennethwright870]

Excellent

[@padraiggluck2980]

There’s a third (interesting) solution. Regard these line segments as vectors, which have magnitude and direction but not location. Translate the red vector three units left and translate the blue vector one unit left and seven units down. The vectors begin at the origin and terminate at (-3,4) and (-1,-7), respectively. The red vector has magnitude 5, the blue vector has magnitude 5*sqrt2) ~= 7.07 units, and angles x and y are contiguous. Take the l2norm of the terminal points, they are 5*sqrt(5) ~= 11.18 units apart. Now use the Law of Cosines, which is easy, to get 135 degrees.

[@Mediterranean81]

tan x = 4/3 tan y=7 tan (x+y) = (tanx + tany)/1-tan x*tan y tan (x+y) = (4/3+7)/1-7*4/3 tan (x+y) = (25/3)/-25/3 tan (x+y) = -1 x+y = arctan (-1) since x and y are acute angles so x+y

[@soniamariadasilveira7003]

I loved your explanations, thanks teacher!

[@PreMath]

I'm so glad! ❤️🌹

[@DB-lg5sq]

tan(x+y)=-1 x+y=135

[@tombufford136]

Thank you , impressive !

[@williamwingo4740]

All angles in degrees: x = arctan(4/3) = 53.13; y = arctan(7) = 81.86; x + y = 53.19 + 81.85 = 134.99 degrees. 🤠

[@PreMath]

Thanks 🌹

[@mathbynisharsir5586]

Unique sir

[@PreMath]

Thanks dear ❤️

[@unknownidentity2846]

Let's have a try: . .. ... .... ..... tan(x+y) = sin(x+y) / cos(x+y) = [sin(x)cos(y) + sin(y)cos(x)] / [cos(x)cos(y) − sin(x)sin(y)] = [sin(x)/cos(x) + sin(y)/cos(y)] / {1 − [sin(x)/cos(x)][sin(y)/cos(y)]} = [tan(x) + tan(y)] / [1 − tan(x)tan(y)] Here we have: tan(x+y) = (4/3 + 7) / (1 − (4/3)*7) = (4 + 21) / (3 − 4*7) = 25 / (3 − 28) = 25 / (−25) = −1 ⇒ x+y = 135°

[@PreMath]

Super! Thanks ❤️

[@MrPaulc222]

I went for tan(-1)(4/3) + tan(-1)(7), which the calculator gave as 135 degrees.

[@PreMath]

thanks ❤️

[@j.r.1210]

Yes, this method is so fast and obvious that I'm surprised it wasn't presented as an option. Indeed, here it is THE option, since this is what trigonometry is for. If the goal is to mess around with rotating triangles, then the instructions should say, "Solve this easy trigonometry problem WITHOUT using trig."

[@abdulkadirbuyuksoy2076]

Woooooow .

[@fikirfikir3039]

( Y=atan7/1= 81.87 ) ... ( x=atan4/3= 53.13 )>>( Y+X=135°)

[@PreMath]

Thanks ❤️🌹

[@fikirfikir3039]

@@PreMath ❤

[@misterenter-iz7rz]

tan x+y=(7+4/3)/(1- 7 4/3)=(25/3)/(1-28/3)=(25/3)/(-25/3)=-1, thus d+y=135 😊

[@PreMath]

Thanks ❤️

[@Ibrahimfamilyvlog2097l]

Very nice 👍❤❤

[@PreMath]

Thank you! Cheers! 🌹

[@giuseppemalaguti435]

x+y=arctg7+arctg4/3=t....tgt=(7+4/3)/(1-7*4/3)=-1....t=135°

[@PreMath]

Thanks ❤️

[@murdock5537]

Nice! Love your first method, Sir, you are great! φ = 30°; x + y > 3φ; sin⁡(x) = 4/5 → cos⁡(x) = 3/5 sin⁡(y) = 7√2/10 → cos⁡(y) = √2/10 → sin⁡(x + y) = sin⁡(x)cos⁡(y) + sin⁡(y)cos⁡(x) = √2/2 → L1 = x + y = 9φ/2 L2 = x + y = √2/2 = 3φ/2 < 3φ ≠ solution or: φ = 30°; ∎ABCD → AB = CD = 5 = AF + BF = 3 + 2 = CG + DG = 4 + 1 = CS + DS = 1 + 4 BC = AD = 7 = AV + DV = 3 + 4 = AE + DE = 4 + 3 → ∆ AFE → EF = 5 shift AG → FS = 5√2 → ∆ FSE → FE = SE = 5 → SAB = y ↔AFE = x → EFS = 3φ/2 → x + y = 9φ/2 ∆ AFE = ∆ CSD → DS = AE = 4 ↔AF = CD = 3 → CS = EF = 5

[@PreMath]

Thanks dear ❤️🌹

[@wackojacko3962]

I trust 2nd method over 1st method. ...but really appreciate both methods for solving. 🙂

[@PreMath]

Great! ❤️

[@amarendrasingh7327]

This is really good question 😊 but I take less than 2 minutes to solve after seeing it Keep uploading more interesting questions videos 😊

[@PreMath]

Keep it up ❤️

[@fsyi8395]

1st method is smarter we don't need trigonometry for special angles

[@padraiggluck2980]

arctan(7)+arctan(4/3) = 135

[@peterkrauliz5400]

That's the one!

[@alster724]

Trigonometry (2nd method) is easier

[@PreMath]

Great ❤️

[@Skank_and_Gutterboy]

Arctangent is your friend.

[@Copernicusfreud]

Yay!, I solved the problem. I calculated the lengths of the sides of all the triangles. My first method was to use the cosine trigonometry functions, cosine x = adjacent/hypotenuse and cosine y = adjacent/hypotenuse and added the numbers. My second method was to use the law of sines to calculate angle x and angle y, and then add those numbers. Both methods gave me the answer in the video.

[@PreMath]

Great job! ❤️