Can you find area of the triangle ABC? | (Law of Cosines) | #math #maths | #geometry

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!

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

Can you find area of the triangle ABC? | (Law of Cosines) | #math #maths | #geometry

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

#FindTriangleArea #LawOfCosines #SimilarTriangles #Triangle #GeometryMath #PythagoreanTheorem #AreaOfTriangle #IsoscelesTriangles
#MathOlympiad #CongruentTriangles
#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.

[RAG]

Excellent method for finding x, but once you have x, at 9 minutes, the height of your triangle is xsin60 = 3rt3/2, so area is 1/2x37x3rt3/2 = 111rt3/4.

[AdemolaAderibigbe-js]

Draw a line from C perpendicular to AD intersecting it at E. Then /_CED = 90 degrees, /_CDE = 60 degrees and /_ECD = 30 degrees. If CE = h, then ED = h/sqrt(3) from properties of a 30-60-90 degree triangle. Let /_CBE = alpha, then /_BCE = (90 - alpha) and /_ACE = 150 -(90- alpha) = 60 + alpha and /_CAE = 90 - (60 + alpha) = 30 - alpha (sum of angles of triangle CAE is 180 degrees).
Next we use the sine rule on triangle CAE as follows:
Sin(60 + alpha)/AE = Sin(30 - alpha)/CE or
Sin(60 + alpha)/(7 - h/sqrt(3)) = Sin(30 - alpha)/h. ----(1)
Then we apply the sine rule to triangle CBE as follows
Sin(alpha)/CE = Sin(90 - alpha)/EB or
Sin(alpha)/h = Sin(90 -alpha)/(30+h/sqrt(3)) ---(2)
Next we expand the Sin angles and substitute the values for SIne and Cosine of angles 30, 60 and 90 degrees and end up with two equations for Tan(alpha) which we combine to get a quadratic equation for h: [h^2 + (201/(2*sqrt(3))*h - 315/2 = 0] which can be solved to give h = 2.59808.
The area of triangle ABC is (1/2)*AB*CE = (1/2)*37*2.59808 = 48.064 (which is the same as (111*sqrt(3))/4).

[yalchingedikgedik]

Good evining Sir
Thanks for your efforts
That’s very nice
With my glades
❤❤❤❤❤

[wackojacko]

So cool! The Pythagorean Theorem is a Special Case of the Law of Cosines. 🙂 Had a lot of fun solving. 😉

[georgexomeritakis]

I solved this by drawing a segment CE vertical to AD and found that ED = 1.5. I used the formula tan(A+B) = (tan A + tan B) / (1 - tan A * tan B), knowing that A + B = 30.

[johnwindisch]

For me it was a teaching moment.
Following the method and overcoming the confusion of the figure being not to scale (😂)
Great problem!

[xualain]

I have an alternative way to find x=3 without having to construct any lines using trigonometry only. Here is my version.
Let Angle CAB=a then angle ACD=120-a, angle ABC=30-a and angle BCD=30+a.
Law of sine for triangle ACD,
sin a/ x=sin(120-a)/7. ……(1)
Similarly for triangle BCD,
sin(30-a)/x=sin(30+a)/30. …..(2)
Dividing (1) by (2)
sin
2*sin
-cos(2*a+30)+cos 30=(-cos(150-2*a)+cos 90)*30/7


cos(2*a+30)=7/74*sqrt(3) from which, we get sin(2*a+30)=73/74
30+sin(2*a+30)*sin 30=47/74
cos(2*a)=2*cos(a)^2-1=47/74 from which tan(a)=3/11*sqrt(3) or cot(a)=11/(3*sqrt(3))
From (1)

[muntasirstudent]

there are total 3 methods to solve it wow