Can you find the angle X? | (Calculators Not Allowed) |

Learn how to find the angle X. Important Geometry, Trigonometry, and Algebra skills are also explained: area of a triangle formula; SOHCAHTOA. 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 the angle...
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 the angle X? | (Calculators Not Allowed) |#math #maths | #geometry
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAngleX#TriangleArea #SOHCAHTOA #Trigonometry #GeometryMath
#MathOlympiad #IntersectingChordsTheorem #RightTriangle #RightTriangles
#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 #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
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
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
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Calculate the length AB
Pythagorean Theorem
Right triangles
Intersecting Chords Theorem
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.

[@Teamstudy4595]

1$t View ❤

[@PreMath]

Thanks dear❤️

[@misterenter-iz7rz]

1250=1/2 ×100 sin x × 100 cos x=2500 sin 2x, sin 2x=1/2, 2x=30° or 150°, x=15° or 75°.😊

[@marcgriselhubert3915]

Fine.

[@PreMath]

Thanks for sharing ❤️

[@Birol731]

yes, we have two values: y₁= 25(√6+√2) x₁= 25*(√6-√2) ⇒ θ₁= 75° y₂= 25(√6-√2) x₂= 25*(√6+√2) ⇒ θ₂= 15°

[@AmirgabYT2185]

My method: Let's build an equivalent triangle under our given triangle. Then a hypotenuse will also be 100 and an angle between these hypotenuses will be 2x. So, we know the sides and an area which is also doubled, so we can find an angle from an equation: 100²sin2x/2=2500 5000sin2x=2500 sin2x=2500/5000=1/2 sin2x=1/2 There are infinite solutions, but in our case angle is acute, so there's one solution for 2x: 2x=30° x=15°

[@jamfocus]

excellent thank you for this

[@libertarianguy5567]

Sin2X=1/2 could also be 150 which would make x=75.

[@krislegends]

​@@libertarianguy5567using the Pythagorean Theorem, which defines a right triangle, proves that 15 degrees is the only solution for x.

[@PreMath]

Excellent Thanks for sharing ❤️

[@phungpham1725]

Thank you! I did it the same way!

[@JLvatron]

Excellent!

[@christianaxel9719]

Notice that ab/2=1250 and a²+b²=100² with tanx=b/a can be transformed to (b/a)/2500=1/a², 1+(b/a)²=100²/a²=100²(b/a)/2500=4(b/a), so tanx²-4tanx+1=0, so tanx=2±√3, and finally tanx=15º or tanx=75º.

[@christianaxel9719]

Algebraic method: if a,b are the other sides of the triangle, ab=1250(2)=2500 and a²+b²=10000, then (a+b)²=10000+2(2500)=15000, (a-b)²=10000-2(2500)=5000, then a+b=50√6, a-b=±50√2; solving equation system: , a=25(√6+√2), b=25(√6-√2) or a=25(√6-√2), b=25(√6+√2), finally tanx=2+√3 or tanx2-√3, and x=75º or x=15º. There are TWO solutions to this problem.

[@alexniklas8777]

From angle B we draw the median BO to the hypotenuse, AO=CO=BO=50; And also the height BD=h. From 1250=(1/2)×100×h, h=2×1250/100=25. BD/BO= sin(2x); sin(2x)=25/50=1/2; 2x=30°; x=15°.

[@prossvay8744]

Let AB=a ; BC==b Area of triangle=1/2ab=1250cm^2 ab=2500 a^2+b^2=100^2 (a+b)^2-2ab=10000 (a+b)^2-5000=10000 (a+b)^2=15000 a+b=50√6 ; a+b=-50√6 (a-b)^2+2ab=10000 a-b=50√2 ; a-b=-50√2 2a=50√6+50√2=50√2(√3+1) a=25√2(√3+1) b=50√6-25√6-25√2 b=25√2(√3-1) Tan(x)=25√2(√3-1)/25√2(√3+1)=(√3-1)/√(√3+1) x=15°

[@PreMath]

Excellent! Thanks for sharing ❤️

[@johnbrennan3372]

Nice method

[@jimlocke9320]

One can make a good educated guess that the right triangle is one of those that appear frequently in problems, so we should try those first, especially after being given the clue that calculators are not allowed. Let A = area and h = hypotenuse. So, we try 45°-45°-90°. A = h²/4 = 10000/4 = 2500. Then we try 30°-60°-90°. A = (h²)(√3)/8. Since there is a radical, the area can not be an integer. Next we try 15°-75°-90°, which appears so often in problems that we are familiar with its properties. A = h²/8 = (100)²/8 = 10000/8 = 1250. We have a match! Our answer is x = 15° or 75°.

[@ramanivenkata3161]

Well explained

[@PreMath]

Glad to hear that! Thanks for the feedback, Ramani dear ❤️

[@murdock5537]

Nice! φ = 30°; ∆ ABC → AC = 100; AB = a; BC = b; CAB = x = ? ab/2 = 1250 → b = 2500/a → √(100^2 - a^2) = 2500/a k ∶= a^2 → k1, k2 = 2500(2 ± √3) → sin⁡(x) = b/100 = (√2/4)(√3 - 1) → x = φ/2

[@LuisdeBritoCamacho]

I did solve the given Problem this way : 1) Divide 100 by 100. You get the hypotenuse equal to 1. h = 1 cm 2) Divide 1.250 * 2 = 2.500 by 100^2. As the Linear Ratio is 100 the second degree Ratio is equal to 100^2 = 10.000 3) Now we have a Triangle, of Hypotenuse = 1 and Area = 0,25 / 2 = 0,125 sq cm 4) Now, if the sides of the Triangle were equal we will have that 50 * 50 / 2 = 2.500 / 2 = 1.250 sq cm. But h^2 = 50^2 + 50^2 ; h^2 = 2.500 + 2.500 ; h^2 = 5.000 ; h = sqrt(5.000) ; h ~ 70,7 5) One must conclude that Length AB different from Length BC. 6) Tan(x) = BC / AB, but we don't know the Length of the Cathetus to find the value of x. 7) What we know is that sin(x) = BC / 100 and the cos(x) = AB / 100. 8) Now, in the Reduced Triangle [ABC] : AB * BC = 0,125 sq cm and AB^2 + BC^2 = 1 9) Doing this fastidious calculations we can achieve our goal, wich is AB = 0,966 cm and BC = 0,259 cm. 10) Multiplying these results by 100 we have : AB = 96,6 cm and BC = 25,9 cm. 11) Check h^2 = 96,6^2 + 25,9^2 ; 100^2 = 9.331,56 + 670,81 100^2 = 10.002,37 wich is a good approximation!! 12) tan(x) = BC / AB ; tan(x) = 25,9 / 96,6 ; tan(x) = 0,268 13) It returns me that x ~ 15º 14) That's all.

[@devondevon4366]

15 That is a 75 15 90 degree triangle in which the hypotenuse square divided by 8 = area Hence 100^2 /8 = 10,000/8 = 1,250 So x = 15 degrees if it faces the smaller side and 75 degrees if it faces the longer sides. It appears it is facing the smaller side , so x =15 degrees if the hypotenuse is 50 , then the area = 50^2/8 =312.5

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@soli9mana-soli4953]

Being the hypotenuse the base we can find its height =1250*2/100=25 Drawing the median from point B, BH = CH = AH = 50 Being the height the half of the median we get a 30,60,90 right triangle in which 30 is the external angle of the isosceles triangle with side equal to median by construction. So x = 15

[@rabotaakk-nw9nm]

👍🥰

[@soli9mana-soli4953]

@@rabotaakk-nw9nm ❤️

[@waheisel]

Again I did this the hard way; I solved for 2 equations (A^2+B^2=100^2 and (A*B)/2=1250 using the quadratic equation and denesting the radical. After a fair amount of algebra you get BC=25(sqrt6-sqrt2) (after defining BC as the short side of the triangle), and AB=25(sqrt6+sqrt2). Construct a 60 degree angle at C that intersects AB at point D. Triangle ABD is 30-60-90. Therefore BD =BC*sqrt3. And CD=BC*2 AD=AB-BD which turns out to be equal to CD. Therefore triangle ADC is isosceles with angle D=150 (supplementary to 30). Therefore angle A=15 Define AB as the short side of the triangle and get angle A=75, the other solution. Another fun puzzle, thanks PreMath!

[@himo3485]

AB=25(√6+√2) BC=25(√6-√2) AC=100 √6+√2 : √6-√2 : 4 = 75° 15° 90° ∠x=15°

[@jamfocus]

learning the proportions of the std triangle 15-75-90 is priceless 👍👍👍

[@mikeparfitt8897]

Please note, the diagram may not be true to scale. Equally valid is AB=25(√6-√2) and BC=25(√6+√2) and ∠x=75°

[@PreMath]

Thanks for sharing ❤️

[@dariosilva85]

The answer could be 15 och 75. ( sin(2x) = 0.5 has two solutions: 2x = 30 + n360 or 2x = (180 - 30) + n360 )

[@alster724]

Double Angle Identity did the trick! A= (ab)/2 2500 = (100)²(sin x)(cos x) 2500= 10000sinxcosx 1= 4 sinxcosx 1= 2(2sinxcosx) 1= 2(sin2x) 1/2= sin 2x 30° = 2x 15° = x

[@christianaxel9719]

Geometric - and fastest - solution: AC is diameter of a circle containing points ABC with center O is at middle of AC with radius r=50. OB is radius too, so OB=r=50. Then AOB is isosceles so ∠ABO=∠OAB=x. Trace height DB from B to AC, then 100DB/2=1250, so DB=25. BDO is a rectangle triangle with hypotenuse OB=r=50 and one cathetus DB=25 then ∠DBO=60º and ∠DOB=30º Depending of position, height DB can be below or above to OB, so

[@CloudBushyMath]

🏋Wonderful Math GYM🏋‍♂🏋‍♀

[@Ihsan403]

👍

[@devondevon4366]

If you divide the hypotenuse square or 100^2 by 8 and get 1250, then it is a 15-75-90 degree, right triangle Why? let's call the two unknown 'a' and 'b' then its area = a * b * 1/2, but we are not given those two sides, only the length of the hypotenuse Since we have two sides and an angle, we can use the law of sine a /sine a = b/sine b = c/sine c Hence a = 100 * sine 15 -------------- sine 90 and b = 100 * sine 75 --------- since 90 Hence, the area of the triangle a * b * 1/2 or 100 * sine 15 100 * sine 75 1 ---------- * ----------- * ----- sine 90 sine 90 2 Let's do a little rearrangement. 100 * 100 * sine 15 * sine 75 1 ---------------------------------------------- * --- sine 90 * sine 90 2 Note that sine 90 degrees = 1 100 ^2 * sine 15 * 75 1 --------------------------------- * ------- 1 * 1 2 note that sine 15 * 75 (degrees) = 0.25 100^2 * 0.25 1 ------------------------' * -- 1 2 as 0.25 =1/4 100^2 * 1/4 * 1/2 100^2 * 1/8 (as 1/4 * 1/2 = 1/8) 100^2 --------- 8 but 100 is the length of the hypotenuse so the hypotenuse square divided by 8 is the area of a 15-75-90 degree right triangle Of course, if you are given the two sides, we multiply them by 1/2 or ( or divide them by 2) To get the area, but what if we are not? PS note a= 25.888 and b= 95.59 and 25.88 x 95.59 = 2500 divided by 2 = 1250 the area

[@hongningsuen1348]

There should be 2 answers for x, x= 15 and x = 75. As sin(2x) = 1/2 => 2x = 30 or 2x = (180 - 30) = 150. This is just a swap of complementary angles of the right-angled triangle. The key to solution of this problem is that for a right-angled triangle with given hypotenuse, the area formulae has the implicit factor of sinxcosx while x is solvable by double angle formulae for sin as sin(2x) = 2cosxsinx.

[@PreMath]

Thanks for the feedback ❤️

[@LuisdeBritoCamacho]

This is a Problem that is easily solve by a System of Two Non Linear Equations: 1) x^y + y^2 = 10.000 2) x * y = 2.500 The solution here proposed (without calculator) is a sort of Reverse Engineering. Trying to rewrite the Object's User Guide knowing its parts. I don't agree with this Vision of Mathematics, that Solving a Problem is doing some sort of Magician Trick!!

[@wackojacko3962]

Doesn't matter,....The top 5% International Mathematical Olympiads can teach Substitution. Therefore they can deal with the logic and determine an unknown given clues like yesterday's puzzle. 🙂

[@PreMath]

Excellent! Thanks for the feedback ❤️

[@dirklutz2818]

Because of symmetry, the angle x=15° or x=75°

[@laxmikantbondre338]

👍Also the answer can be 75. Sin2x = 1/2. So 2X = 150. So x = 75

[@AmirgabYT2185]

x=15°

[@almosawymehdi3416]

x can be equal to 75° or not? Because I have found two solution, 15° and 75°

[@krislegends]

With sin(2x) = 1/2, the solution to x can be infinite. Using the Pythagorean Theorem, I get 15 degrees.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@Birol731]

My way of solution is ➡ 1250 cm²= AB*BC/2 2500= AB*BC AB= x BC= y ⇒ x*y=2500 we can also write: (x+y)²= x²+y²+2xy x²+y²= c² c= 100 c²= 10.000 ⇒ (x+y)²= 10.000+2*2500 (x+y)²= 15.000 x+y= √15.000 x+y= 50√6 xy= 2500 x= 2500/y ⇒ (2500/y)+y= 50√6 2500+y²= 50√6 y y²-50√6y+2500=0 Δ= b²-4ac Δ= 15.000-10.000 Δ= 5000 √Δ= 50√2 y₁= (50√6+50√2)/2 y₁= 25(√6+√2) y₂= (50√6-50√2)/2 y₂= 25(√6-√2) x₁= 2500/y₁ x₁= 2500/25(√6+√2) x₁= 100/(√6+√2) x₁= 100*(√6-√2)/(√6²-√2²) x₁= 100*(√6-√2)/(6-2) x₁= 25*(√6-√2) x₂= 2500/y₂ x₂= 2500/ 25(√6-√2) x₂= 100/(√6-√2) x₂= 100*(√6+√2)/(√6²-√2²) x₂= 100*(√6+√2)/(6-2) x₂= 25*(√6+√2) tan(θ₁)= y₁/x₁ tan(θ₁)= 25(√6+√2)/25(√6-√2) tan(θ₁)= (√6+√2)/(√6-√2) tan(θ₁)= (√6+√2)²/(√6²-√2²) tan(θ₁)= (6+2√12+2)/4 tan(θ₁)= (8+4√3)/4 θ₁= arctan(2+√3) θ₁= 75° tan(θ₂)= y₂/x₂ tan(θ₂)= 25(√6-√2)/25(√6+√2) tan(θ₂)= (√6-√2)/(√6+√2) tan(θ₂)= (√6-√2)²/(√6²-√2²) tan(θ₂)= (6-2√12+2)/4 tan(θ₂)= (8-4√3)/4 θ₂= arctan(2-√3) θ₂= 15° so we have here 2 values for the angle x or θ value: y₁= 25(√6+√2) x₁= 25*(√6-√2) ⇒ θ₁= 75° y₂= 25(√6-√2) x₂= 25*(√6+√2) ⇒ θ₂= 15°

[@MegaSuperEnrique]

Sin(2x)=1/2 so 2x = 150 or 30, x= 75 or 15

[@PreMath]

Thanks for sharing ❤️

[@quigonkenny]

Let AB = a and BC = b. A = bh/2 1250 = ab/2 ab = 2(1250) = 2500 b = 2500/a a² + b² = c² a² + (2500/a)² = 100² a² + 6250000/a² = 10000 a⁴ + 6250000 = 10000a² a⁴ - 10⁴a² + 625•10⁴ = 0 a² = -(-10⁴)±√(-10⁴)²-4(1)(625•10⁴) / 2(1) a² = 5000 ± (√10⁸-2500•10⁴)/2 a² = 5000 ± √75•10⁶/2 a² = 5000 ± 5000√3/2 = 5000 ± 2500√3 a² = 2500(2±√3) a = 50√(2±√3) a₁ = 50√(2+√3) | a₂ = 50√(2-√3) b₁ = 2500/50√(2+√3) b₁ = 50/√(2+√3) b₁ = 50√(2-√3)/√(2+√3)√(2-√3) b₁ = 50√(2-√3)/√(4-3) = 50√(2-√3) b₂ = 2500/50√(2-√3) b₂ = 50/√(2-√3) b₂ = 50√(2+√3)/√(2-√3)√(2+√3) b₂ = 50√(2+√3)/√(4-3) = 50√(2+√3) Will go with a₁ and b₁ as a > b in the figure as drawn. sin(x) = b/100 = 50√(2-√3)/100 sin(x) = √(2-√3)/2 = √(4-2√3)/2√2 sin(x) = √(3-2√3+1)/2√2 sin(x) = √(√3-1)²/2√2 sin(x) = (√3-1)/2√2 = √3/2√2 - 1/2√2 sin(x) = (√3/2)(1/√2) - (1/2)(1/√2) sin(x) = sin(60°)cos(45°) - cos(60°)sin(45°) sin(x) = sin(60°-45°) = sin(15°) x = 15° x = 75° is also a solution (using a₂, b above) assuming b > a.

[@adept7474]

ВН ⟂ АС. ВН = 25. Сircle on АС (О - сenter). АО = ВО = 50. In ▲ОНВ: ВН/ВО = 1/2. ∠ВОН = 30°, х = 15°

[@PreMath]

Excellent! Thanks for sharing ❤️

[@user-uv6cc6pv7f]

x=15°,75°

[@PreMath]

Thanks for sharing ❤️

[@mathematics_.T.]

Pitago

[@PrithwirajSen-nj6qq]

AB =a BC=b CA =c a^2+ b^2=10000 1/2absinB =1250 ab=1250*2/sinB=2500 (a+b)^2=12500 (a-b) ^2=10000-2500=7500 From here a and b will be known. Sin x = b/c x=sin inverse b/c

[@PreMath]

Thanks for sharing ❤️