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

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29 күн бұрын

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
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Пікірлер: 63

@Teamstudy4595 27 күн бұрын

1$t View ❤

@PreMath 27 күн бұрын

Thanks dear❤️

@AmirgabYT2185 27 күн бұрын

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 27 күн бұрын

excellent thank you for this

@libertarianguy5567 27 күн бұрын

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

@krislegends 27 күн бұрын

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

@PreMath 27 күн бұрын

Excellent Thanks for sharing ❤️

@phungpham1725 27 күн бұрын

Thank you! I did it the same way!

@misterenter-iz7rz 27 күн бұрын

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 27 күн бұрын

Fine.

@PreMath 27 күн бұрын

Thanks for sharing ❤️

@Birol731 27 күн бұрын

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°

@JLvatron 27 күн бұрын

Excellent!

@alexniklas8777 27 күн бұрын

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°.

@christianaxel9719 27 күн бұрын

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 27 күн бұрын

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.

@jimlocke9320 27 күн бұрын

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 27 күн бұрын

Well explained

@PreMath 27 күн бұрын

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

@jamestalbott4499 27 күн бұрын

Thank you!

@PreMath 27 күн бұрын

You are very welcome! Thanks ❤️

@prossvay8744 27 күн бұрын

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 27 күн бұрын

Excellent! Thanks for sharing ❤️

@johnbrennan3372 27 күн бұрын

Nice method

@soli9mana-soli4953 27 күн бұрын

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 20 күн бұрын

👍🥰

@soli9mana-soli4953 16 күн бұрын

@@rabotaakk-nw9nm ❤️

@devondevon4366 27 күн бұрын

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

@murdock5537 27 күн бұрын

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

@christianaxel9719 27 күн бұрын

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

@waheisel 20 күн бұрын

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!

@Ihsan403 27 күн бұрын

👍

@LuisdeBritoCamacho 27 күн бұрын

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.

@dariosilva85 27 күн бұрын

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

@CloudBushyMath 27 күн бұрын

🏋Wonderful Math GYM🏋‍♂🏋‍♀

@alster724 27 күн бұрын

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

@hongningsuen1348 27 күн бұрын

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 27 күн бұрын

Thanks for the feedback ❤️

@wackojacko3962 27 күн бұрын

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 27 күн бұрын

Excellent! Thanks for the feedback ❤️

@dirklutz2818 27 күн бұрын

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

@devondevon4366 26 күн бұрын

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

@almosawymehdi3416 25 күн бұрын

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

@LuisdeBritoCamacho 26 күн бұрын

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!!

@himo3485 27 күн бұрын

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

@jamfocus 27 күн бұрын

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

@mikeparfitt8897 27 күн бұрын

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 27 күн бұрын

Thanks for sharing ❤️

@laxmikantbondre338 27 күн бұрын

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

@adept7474 27 күн бұрын

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

@PreMath 27 күн бұрын

Excellent! Thanks for sharing ❤️

@krislegends 27 күн бұрын

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

@PreMath 27 күн бұрын

Excellent! Thanks for sharing ❤️

@Birol731 27 күн бұрын

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°

@AmirgabYT2185 27 күн бұрын

x=15°

@MegaSuperEnrique 27 күн бұрын

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

@PreMath 27 күн бұрын

Thanks for sharing ❤️

@quigonkenny 27 күн бұрын

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.

@mathematics_.T. 24 күн бұрын

Pitago

@user-uv6cc6pv7f 27 күн бұрын

x=15°,75°

@PreMath 27 күн бұрын

Thanks for sharing ❤️

@PrithwirajSen-nj6qq 27 күн бұрын

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 27 күн бұрын

Thanks for sharing ❤️