Can you find angle X and Y values? | (Law of Sines) |

Learn how to find the angle X and Y values. Important Geometry skills are also explained: Law of Sines; Trigonometry. Step-by-step tutorial by PreMath.com
Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!
Step-by-step tutorial by PreMath.com
• Can you find angle X a...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Can you find angle X and Y values? | (Law of Sines) | #math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAngleX #FindAngleY #LawOfSines #Trigonometry #Triangle #GeometryMath #PythagoreanTheorem #AreaOfTriangle #IsoscelesTriangles
#MathOlympiad #CongruentTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #ExteriorAngleTheorem #PythagoreanTheorem #IsoscelesTriangles #AreaOfTriangleFormula #AuxiliaryLines
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #CompetitiveExams #CompetitiveExam
#MathematicalOlympiad #OlympiadMathematics #LinearFunction #TrigonometricRatios
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
Andy Math
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Olympiad Question
Find Area of the Shaded Triangle
Geometry
Geometry math
Geometry skills
Right triangles
Exterior Angle Theorem
pythagorean theorem
Isosceles Triangles
Area of the Triangle Formula
Competitive exams
Competitive exam
Find Area of the Triangle without Trigonometry
Find Area of the Triangle
Auxiliary lines method
Pythagorean Theorem
Square
Diagonal
Congruent Triangles
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

[@bkp_s]

Nice video sir

[@PreMath]

Glad to hear that! Thanks for the feedback ❤️

[@soli9mana-soli4953]

Being ABC isosceles and the sum of angles ADB + ADC = 240° (and setting ADB=alpha, ADC = 240 - alpha) we can write the sines law: 1) AD/sin 50° = AB/sin alpha 2) AD/sin 30° = AC/sin (240 - alpha) solving both identities by AD and comparing we get: sin alpha/sin 50° = sin (240 - alpha)/sin 30° and finally we have: tan alpha = sqrt 3*sin 50°/sin 50° - 1 alpha = - 80° => alpha = 100 so in ADB x = 180 - 50 - 100 = 30 being x + y = 40 => y = 40 - x = 40 - 30 = 10

[@georgexomeritakis2793]

You can solve this problem with Euclidean geometry. Construct a trapezoid BCDE with base angles equal to 40. Then it's easy to prove that segments BE=ED=DC. Now triangle AED is isosceles due to symmetry. Also if you extent BD to intersect AC at point F, we create a right triangle CDF with angles 30,60,90. Now segment DF is half of CD. Now compare isosceles triangle AED and right triangle ADF. It is obvious that ADF is half of AED so angle DAF is half of angle EAD.

[@robertwarburton9944]

Angle ADF is equal to AED. Note if G is the mid point of ED then triangles AGD and ADF are congruent. Therefore angle GAD = DAF = y. By symmetry EAD = 2y and also by symmetry BAE = y. Therefore 4y = 40, hence y = 10 and so x = 30.

[@afjalhussainlaskar7352]

Thank you sir

[@jamestalbott4499]

Thank you!

[@michaeldoerr5810]

That is probably the most useful application of a double angle identity. I shall use that for practice!!!

[@SGuerra]

Parabéns pela questão! Ela é muito boa! Brasil - Outubro de 2024.

[@rey-dq3nx]

Beautiful

[@SumitVerma-lg3qh]

End was nice 1/2 ----> sin 30

[@waheisel]

I didn't come close to this trigonometric solution, nor to a geometric solution. So I resorted to a Cartesian coordinate and a trig calculator: set the origin at B and (2,0) at C. Midpoint of BC, E is (1,0). A will be (1,arctan70) or (1,2.7475). Any number with a decimal is approximate. line BD is y=tan20x=0.364x. Line CD is y=-tan40(x-2)=-0.839x+1.678. Solve for intersection of those lines, i.e. coordinates of D and you get (1.335,0.486). Using the coordinates of A and D you get angle EAD=arctan 0.395/2.240=10. Since EAD+y is 20 (half of BAC=40), y=10 and x=30.

[@jimlocke9320]

Note that

[@wackojacko3962]

Sine Sine Everywhere a Sine Blockin out the scenery Breakin my mind Do this, don't do that Can't you read the Sine? 🙂

[@montynorth3009]

Let length BC = 1 for simplicity. In triangle BDC. BC /sin 120 = BD / sin 40. BD = 0.7422. In triangle ABC. AB / sin 70 = 1 / sin 40. AB = 1.4619. In triangle BDA. Cosine Rule. AD^2 = 1.4619^2 + 0.7422^2 - 2 * 1.4619 * 0.7422 * cos 50. AD = 1.1372. 1.1372 /sin 50 = 0.7422 / sin x. sin x = 0.49996, allowing for rounding errors = 0.5. x = 30 degrees. x + y = 40 from triangle abc, so y = 10 degrees.

[@deniseockey6204]

This is also the way I did this problem. Is is A LOT BETTER than the endless trig function manipulation. I did not like this problem at all.

[@sanjeebrana4209]

x=30°,y=10° and x+y=40°

[@축복이-x6u]

asnwer= 55 & 25 isit

[@축복이-x6u]

answer=10& 30 isit

[@xualain3129]

I got a simpler solution to share with. First we have x+y=180-50-20-30-40=40 and AB=AC y=40-x sin(50+x)/sin 50=sin(30+y)/sin 30 ->sin(50+x)=2*sin 50*sin(30+40-x) using sin 30=1/2 sin(50+x)=2*sin 50*sin(70-x)=2*cos 40*sin(70-x)=sin(110-x)+sin(30-x) -> sin(50+x)-sin(110-x)=sin(30-x) ->sin(x+50)+sin(x-110)=sin(30-x) -> 2*sin(x-30)*cos 80=sin(30-x) ->sin(x-30)(2*cos 80+1)=0 since 2*cos 80+10 hence sin(x-30)=0 we get x=30 and then y=10

[@krishnamacharykothandarama3412]

Isosceles triangle math

[@joeschmo622]

Thar 80/20 to 30/10 is the weirdest thing I've seen in a while. Gonna have to chew on that for a while...

[@Birol731]

My way of solution ▶ By considering the first triangle ΔABD [AB]= [AC] while ∠ABC= ∠BCA= 70° [AB]= a [BD]= d [AD]= c ∠DAB = x ∠ABD = 50° ∠BDA = 130°-x by writing the sinus theorem for the first triangle we get: sin(x)/d= sin(50°)/c= sin(130°-x)/a ⇒ a/c= sin(130°-x)/sin(50°).............Eq-1 b) By considering the second triangle ΔADC [AC]= a [AD]= c [DC]= e ∠CAD = y ∠DCA = 30° ∠ADC = 150°-y by writing the sinus theorem we get: sin(y)/e= sin(30°)/c= sin(150°-y)/a we know that x+y= 40° y= 40°-x ⇒ sin(y)/e= sin(30°)/c= sin(150°-(40-x))/a sin(y)/e= sin(30°)/c= sin(110°+x)/a ⇒ a/c= sin(110°+x)/sin(30°)..............Eq-2 c) By considering the third triangle ΔBCD [DB]= d [CD]= e [BC]= b ∠DBC = 20° ∠BCD = 40° ∠CDB = 120° by writing the sinus theorem we get: sin(120°)/b= sin(20°)/e= sin(40°)/d here we don't have 2 same parameters (like a/c) that we can insert into equation-1 or equation-2........ d) both a/c ratios must be equal to each other, so: sin(130°-x)/sin(50°) = sin(110°+x)/sin(30°) sin(130-x)= sin(130°)*cos(x) - sin(x)*cos(130°) sin(110°+x)= sin(110°)*cos(x) + sin(x)*cos(110°) ⇒ [sin(130°)*cos(x) - sin(x)*cos(130°)]/sin(50°) = [sin(110°)*cos(x) + sin(x)*cos(110°)]/sin(30°) ⇒ sin(30°)*[sin(130°)*cos(x) - sin(x)*cos(130°)]= sin(50°)*[sin(110°)*cos(x) + sin(x)*cos(110°)] ⇒ sin(30°)*sin(130°)*cos(x) - sin(30°)*sin(x)*cos(130°) = sin(50°)*sin(110°)*cos(x) + sin(50°)*sin(x)*cos(110°) ⇒ cos(x)*sin(30°)*sin(130°) - sin(50°)*sin(110°)*cos(x) = sin(30°)*sin(x)*cos(130°) + sin(50°)*sin(x)*cos(110°) ⇒ cos(x)[sin(30°)*sin(130°) - sin(50°)*sin(110°)]= sin(x)[ sin(30°)*cos(130°) + sin(50°)*cos(110°)] ⇒ cos(x)/sin(x)= [ sin(30°)*cos(130°) + sin(50°)*cos(110°)]/[sin(30°)*sin(130°) - sin(50°)*sin(110°)] ⇒ sin(30°)= 0,5 cos(130°)= -0,642787609 sin(50°)= 0,766044443 cos(110°)= - 0,342020143 sin(130°)= 0,766044443 sin(110°)= 0,93969262 ⇒ cot(x)= [ - 0,5*0,642787609 - 0,766044443*0,342020143] / [0,5*0,766044443 - 0,766044443*0,93969262] cot(x)= -0,583396435 / -0,336824088 cot(x)= 1,732050812 cot(x)= √3 arccot(x)= arccot(√3) ⇒ x= 30° y= 10°

[@shaozheang5528]

I created a 30 60 90 triangle

[@kyoshi1950]

There's an easier solution. Because this is a blablabla triangle. We use the complementaries. If we have 180° for all angles of triangle. We have 140° for both. So we lack 40°. How to find for the two? We use the bottom triangle. We get 180-60=120. We consider that D has 120° on all its angles. For BAD = 120+50)-180 we get 10. For CAD = 120+30)-180 we get 30. Because it is a triangle with two angles equal (Isosceles I think?), we swap them. Then we get the answers. Giving... x+y = 40 (if x=10, and y=30) x: y=40-10 = 30 y: x=40-30 = 10 Have a nice day or night.

[@kalkulol]

how did u even put "(120+50)-180 we get 10" while 120 is D and it's unknown if the intersection are set the same degree?

[@giuseppemalaguti435]

AO=a,BO=b,CO=c..x+y=40...a/sin50=b/sinx..a/sin30=c/Siny..inoltre b/sin40=c/sin20..b=2ccos20..risulta sinx/Siny=2cos20sin50/sin30,che unita a x+y=40 da..ctgx=√3....x=30..

[@bobbyheffley4955]

Angle x+angle y=40°

[@JSSTyger]

I randomly chose x=20 and got y=20

[@KipIngram]

This one isn't any fun, man. Just a nasty technical grind - no "clever thinking" involved. I like the ones where recognizing the clever solution approach is the bulk of the work.

[@xualain3129]

I got a less nasty solution above. Try to see if it interests you or not. My best regards.

[@dianewhittington9987]

why not give a clever solution?

[@duobrothersgaming3001]

Who is from class 9th here

[@unknownidentity2846]

Let's face this challenge: . .. ... .... ..... ∠ABC = ∠ABD + ∠CBD = 50° + 20° = 70° ∠ACB = ∠ACD + ∠BCD = 30° + 40° = 70° ⇒ ∠ABC = ∠ACB ⇒ AB = AC From the interior angle sums of the triangles ABC, ABD and ACD we can conclude: x + y = ∠BAD + ∠CAD = ∠BAC = 180° − ∠ABC − ∠ACB = 180° − 70° − 70° = 40° ∠ADB = 180° − ∠ABD − ∠BAD = 180° − 50° − x = 130° − x ∠ADC = 180° − ∠ACD − ∠CAD = 180° − 30° − y = 150° − y Now we apply the law of sines to the triangles ABD and ACD: sin(∠ABD)/AD = sin(∠ADB)/AB sin(∠ACD)/AD = sin(∠ADC)/AC sin(50°)/AD = sin(130° − x)/AB sin(30°)/AD = sin(150° − y)/AC With x=40°−y ⇒ 130°−x=130°−(40°−y)=130°−40°+y=90°+y and AB=AC we obtain: sin(50°)/sin(30°) = sin(90° + y)/sin(150° − y) sin(50°)*sin(150° − y) = sin(30°)*sin(90° + y) sin(50°)*[sin(150°)cos(y) − cos(150°)sin(y)] = (1/2)*[sin(90°)cos(y) + cos(90°)sin(y)] sin(50°)*[(1/2)cos(y) + (√3/2)sin(y)] = (1/2)*cos(y) sin(50°)*[cos(y) + √3*sin(y)] = cos(y) sin(50°)*[1 + √3*tan(y)] = 1 1 + √3*tan(y) = 1/sin(50°) √3*tan(y) = 1/sin(50°) − 1 tan(y) = [1/sin(50°) − 1]/√3 ⇒ y = arctan{[1/sin(50°) − 1]/√3} = 10° ⇒ x = 40 − y = 30° Let's check the result: sin(50°)/sin(30°) = sin(90° + y)/sin(150° − y) sin(50°)/(1/2) = sin(100°)/sin(140°) 2*sin(50°) = sin(100°)/sin(140°) 2*sin(50°)*sin(140°) = sin(100°) 2*sin(50°)*sin(140°) = 2*sin(50°)*cos(50°) sin(140°) = cos(50°) sin(90° + 50°) = cos(50°) cos(50°) = cos(50°) ✓ Best regards from Germany

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال f(x)=sin(40-x)/ sinx 0

[@NishagowdamsNishagowdams]

X=40 and y= 60 so please respond to my feedback

[@PushkarJha44]

It's absolutely wromg

[@JamesDavy2009]

The Triangle Sum formula gave us the equation x + y = 40°. Your answers do not satisfy it.

[@NishagowdamsNishagowdams]

Thanks ​ but I tried it next time it will be perfect@@PushkarJha44

[@NishagowdamsNishagowdams]

​@@JamesDavy2009thanks for rectifying my mistake next time I will be perfect