Can you find the angle X? | (Calculators Not Allowed) |
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@Teamstudy4595 24 күн бұрын
1$t View ❤
@PreMath 24 күн бұрын
Thanks dear❤️
@misterenter-iz7rz 24 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 24 күн бұрын
Fine.
@PreMath 24 күн бұрын
Thanks for sharing ❤️
@Birol731 24 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 24 күн бұрын
Excellent!
@AmirgabYT2185 24 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 24 күн бұрын
excellent thank you for this
@libertarianguy5567 24 күн бұрын
Sin2X=1/2 could also be 150 which would make x=75.
@krislegends 24 күн бұрын
@@libertarianguy5567using the Pythagorean Theorem, which defines a right triangle, proves that 15 degrees is the only solution for x.
@PreMath 24 күн бұрын
Excellent Thanks for sharing ❤️
@phungpham1725 24 күн бұрын
Thank you! I did it the same way!
@alexniklas8777 24 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 23 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.
@ramanivenkata3161 24 күн бұрын
Well explained
@PreMath 24 күн бұрын
Glad to hear that! Thanks for the feedback, Ramani dear ❤️
@jimlocke9320 24 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°.
@christianaxel9719 23 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.
@jamestalbott4499 24 күн бұрын
Thank you!
@PreMath 24 күн бұрын
You are very welcome! Thanks ❤️
@prossvay8744 24 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 24 күн бұрын
Excellent! Thanks for sharing ❤️
@johnbrennan3372 24 күн бұрын
Nice method
@Ihsan403 24 күн бұрын
👍
@soli9mana-soli4953 23 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 16 күн бұрын
👍🥰
@soli9mana-soli4953 12 күн бұрын
@@rabotaakk-nw9nm ❤️
@christianaxel9719 23 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
@devondevon4366 24 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
@dariosilva85 24 күн бұрын
The answer could be 15 och 75. ( sin(2x) = 0.5 has two solutions: 2x = 30 + n360 or 2x = (180 - 30) + n360 )
@waheisel 16 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!
@almosawymehdi3416 22 күн бұрын
x can be equal to 75° or not? Because I have found two solution, 15° and 75°
@hongningsuen1348 24 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 24 күн бұрын
Thanks for the feedback ❤️
@murdock5537 24 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
@LuisdeBritoCamacho 24 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.
@CloudBushyMath 23 күн бұрын
🏋Wonderful Math GYM🏋♂🏋♀
@alster724 23 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
@himo3485 24 күн бұрын
AB=25(√6+√2) BC=25(√6-√2) AC=100 √6+√2 : √6-√2 : 4 = 75° 15° 90° ∠x=15°
@jamfocus 24 күн бұрын
learning the proportions of the std triangle 15-75-90 is priceless 👍👍👍
@mikeparfitt8897 24 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 24 күн бұрын
Thanks for sharing ❤️
@dirklutz2818 23 күн бұрын
Because of symmetry, the angle x=15° or x=75°
@devondevon4366 22 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
@wackojacko3962 24 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 24 күн бұрын
Excellent! Thanks for the feedback ❤️
@LuisdeBritoCamacho 23 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!!
@laxmikantbondre338 24 күн бұрын
👍Also the answer can be 75. Sin2x = 1/2. So 2X = 150. So x = 75
@adept7474 24 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 24 күн бұрын
Excellent! Thanks for sharing ❤️
@MegaSuperEnrique 24 күн бұрын
Sin(2x)=1/2 so 2x = 150 or 30, x= 75 or 15
@PreMath 24 күн бұрын
Thanks for sharing ❤️
@AmirgabYT2185 24 күн бұрын
x=15°
@krislegends 24 күн бұрын
With sin(2x) = 1/2, the solution to x can be infinite. Using the Pythagorean Theorem, I get 15 degrees.
@PreMath 24 күн бұрын
Excellent! Thanks for sharing ❤️
@Birol731 24 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°
@quigonkenny 24 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.
@user-uv6cc6pv7f 24 күн бұрын
x=15°,75°
@PreMath 24 күн бұрын
Thanks for sharing ❤️
@mathematics_.T. 21 күн бұрын
Pitago
@PrithwirajSen-nj6qq 24 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 24 күн бұрын
Thanks for sharing ❤️
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