The problem, as stated, is not valid. When sin(x) or cos(x) is computed, x must be an angle. Angles are typically expressed in degrees or radians. When the angle is a central angle of a circle, radians is the ratio of arc length to radius. Radians, therefore, does not have a dimension. Degrees is radians times 180/π. So it does not have a dimension, either. The problem is not valid because x has a dimension (some unit of length) and sin() and cos() can only be computed for dimensionless quantities. The problem could be made valid by stating that the sin() and cos() are computed for x divided by its unit of length.
Sin(x) and cos(x) have a definition for x real. Functions sin and cos are defined on R and have 2.Pi for period. For x real, sin(x) = sin(x rad, mod 2.Pi), and the same with cos(x). Ex: sin(100) = sin(100 rad) = sin((100 - 32.Pi) rad) = sin(-0.5309 rad) (about) = -0.5063 (about). Radians shoud always be used as angle unit, the use of degrees or other units is the best way to confusions.
@@marcgriselhubert3915 We measure distances in units of measure, such as meters, feet, miles, etc. In geometry problems, we may use the term "units" for the general case. Units could refer to any of these. If the sides of a right triangle have lengths 3 and 4 units, the hypotenuse must have a length of 5 units. The unit can be meters, feet, miles, etc. Radians and degrees refer to ratios. The ratio of the circumference of a circle to its radius is 2π. We say 2π radians but radian is not a unit of measure. It is called a dimensionless quantity, which students often have trouble understanding, especially when there are conversions between radians and degrees. My point is that x is a length and, therefore, must have a dimension. However, angles are measured in dimensionless quantities. You can't compute the sine of 2 meters, for example. On the subject of degrees vs. radians, I do have a calculator that computes the trig functions for angles in degrees. On the other hand, the computer language python has a math function which computes trig functions for angles in radians. (They do have a math function for pi, which provides a simple way of writing a function for the conversion.) I have always seen the special right triangles in geometry, 30°-60°-90° and 45°-45°-90°, given in degrees, not radians.
@@jimlocke9320 If you want to be coherent, x here has no unit, it is simply a positive real number, it is the ratio of the length of AE in a certain unit divided by this unit. this unit can be meters, millimeters, miles, ... That's the reason why you can draw the triangle as big or as small as you want on your paper. Considering degrees, if you were a french student using degrees you would get a bad mark. But I know that in anglo-american countries it is still used and I also often use degrees here when I try to "speak" english (when there is no special problem with that unit).
@@marcgriselhubert3915 Does that mean that the special 30°-60°-90° right triangle is designated π/6 - π/3 - π/2 and the special 45°-45°-90° isosceles right triangle is designated π/4 - π/4 - π/2 in France?
SCOOP: In fact, this problem is physicaly impossible. A right triangle ABC with hypotenuse [A,C] divided by points E and D in 3 equal lengthes represented by the real x, with a = EB = sin(x) and b = DB = cos(x) doen't exist, it is not possible to construct it. If this triangle was possible to construct, then we proved that necessary x = sqrt(5)/5, and then a = sin(x) = 0.4324 (about) and b = cos(x) = 0,9016 (about) Then B is at the intersection of the circle of center E and of radius a with the circle of center D and of radius b. a+b = 1.334, abs(a-b) = 0.4692 and DE = sqrt(5)/5 = 0.447. We do not have abs(a-b) < DE< a+b, so these two circles do not intersect, the construction is impossible. (It is also possible to verify that the circle of diameter [A,C], normaly containing B, does not intersect with the two precedent circles) So there was really a problem with this exercice, but not with sin and cos of a real.
We can use the cosine rule for triangles. So let AB be h and the angle EAB be theta. Then cos(theta) = h/(3x). Applying the cosine rule for triangle AEB yields a^2 = x^2 + h^2 - (2xh)cos(theta) and for triangle ABD gives us b^2 = (2x)^2 + h^2 - (4xh)cos(theta). Substituting a = sin(x) and b = cos(x) and cos(theta) = h/(3x) gives us the two equations: sin(x)^2 = x^2 + h^2 - (2/3)h^2 and cos(x)^2 = 4(x^2) + h^2 - (4/3)h^2. Next we add these two equations giving us 1 = 5(x^2) yielding x = sqrt(5)/5.
We use an orthonormal center B and first axis (BC), we have B(0; 0) A(0; a') C(c; 0) E(c/3; (2/3).a') D((2/3).c; a'/3) with a' = BA and c = BC BE^2 = (1/9).(c^2 +4.(a'^2)) = (sin(x))^2 and BD^2 = (1/9).(a'^2 +4.(c^2)) = (cos(x))^2. As (sin(x))^2 + (cos(x))^2 = 1, we have (1/9)(5.(a'^2) +5.(c^2)) = 1 Then a'^2 + c^2 =9/5. As a'^2 + c^2 = BA^2 + BC^2 = AC^2, we the have that AC^2 = 9/5 and AC = 3/sqrt(5). Finally x = AC/3 = 1/sqrt(5) = (sqrt(5))/5.
a^2+b^2=1 by definition. Therefore triangle EDB must be right triangle with hypotenuse 1. Therefore x=1. And angle EBD must be right angle, but that’s contradict to the conditions
When we have a triangle with sides a,b,c we can generate a formula for the length of a median. Expand that median to generate a parallelogram and we know that the sum of the squares of the sides equels the sum of the squares of the diagonals. The median BN in triangle ABC is 1.5x . And now, in triangle EBD apply the formula.
As someone has pointed out, the answer does not satisfy the triangular inequality. In other words, I solved it using cosine rule and got 1/sqrt(5), but also found that cos(A) is not a real number, its square being equal to -0.0216
x=sqrt(0.2)=.4472 approx a = sin x, so x= inv sin a, meaning a = 26.57 degrees or .4636 radians b = cos x, so x= inv cos b, meaning b= 63.43 degrees or 1.1071 radians a - b is greater than x, whether x is in radians or degrees, meaning the triangle EDB cannot be constructed. Clearly a mistake in setting the question. The mistake is using x as both the angle for calculating a and b, and for the dimensions of AE, ED and DC. They are unrelated quantities.
x is a (positive) real. Sin(x) and cos(x) have a meaning for x real. For x real, sin(x) = sin(angle (x rad mod 2.Pi)), id for cos. Normally we should always use radians and never degrees or other angle units to avoid this type of question.
Let's find x: . .. ... .... ..... First of all we add the points F and G on BC such that BEF and BDG are right triangles. Since ABC is also a right triangle, we can apply the Pythagorean theorem to all these triangles: AC² = AB² + BC² BE² = EF² + BF² BD² = DG² + BG² (3x)² = AB² + BC² sin²(x) = (2*AB/3)² + (BC/3)² cos²(x) = (AB/3)² + (2*BC/3)² 9x² = AB² + BC² sin²(x) = 4*AB²/9 + BC²/9 cos²(x) = AB²/9 + 4*BC²/9 9x² = AB² + BC² sin²(x) + cos²(x) = 4*AB²/9 + BC²/9 + AB²/9 + 4*BC²/9 1 = 5*AB²/9 + 5*BC²/9 = 5*(AB² + BC²)/9 = 5*(9x²)/9 = 5*x² ⇒ x = 1/√5 Best regards from Germany
Hi, Do you have any idea of the value of the smallest angle ACB? (my first guess is: 18 degrees or pi/10 rad) But now I don't think it's possible to calculate the angle! I got n² + m² = 0.2 and n² - m² = cos(2x)/3 = 0.208655 and therefore m² = -0.1043
O segmento a é perpendicular ao segmento AD, visto que divide AD em duas partes iguais, logo, AB = a. Então temos dois triângulos retângulos. Por Pitágoras, igualamos as duas equações e chegamos ao resultado posto pelo professor. Linda questão.
No, there is only one solution: (sqrt(5))/5, as obtained in the proof. Imagine k = 1,000,000 for example. The ABC triangle will be enormous but BE and BD always between 0 and 1... !
The problem, as stated, is not valid. When sin(x) or cos(x) is computed, x must be an angle. Angles are typically expressed in degrees or radians. When the angle is a central angle of a circle, radians is the ratio of arc length to radius. Radians, therefore, does not have a dimension. Degrees is radians times 180/π. So it does not have a dimension, either. The problem is not valid because x has a dimension (some unit of length) and sin() and cos() can only be computed for dimensionless quantities. The problem could be made valid by stating that the sin() and cos() are computed for x divided by its unit of length.
Sin(x) and cos(x) have a definition for x real. Functions sin and cos are defined on R and have 2.Pi for period.
For x real, sin(x) = sin(x rad, mod 2.Pi), and the same with cos(x).
Ex: sin(100) = sin(100 rad) = sin((100 - 32.Pi) rad) = sin(-0.5309 rad) (about) = -0.5063 (about).
Radians shoud always be used as angle unit, the use of degrees or other units is the best way to confusions.
@@marcgriselhubert3915 We measure distances in units of measure, such as meters, feet, miles, etc. In geometry problems, we may use the term "units" for the general case. Units could refer to any of these. If the sides of a right triangle have lengths 3 and 4 units, the hypotenuse must have a length of 5 units. The unit can be meters, feet, miles, etc. Radians and degrees refer to ratios. The ratio of the circumference of a circle to its radius is 2π. We say 2π radians but radian is not a unit of measure. It is called a dimensionless quantity, which students often have trouble understanding, especially when there are conversions between radians and degrees. My point is that x is a length and, therefore, must have a dimension. However, angles are measured in dimensionless quantities. You can't compute the sine of 2 meters, for example.
On the subject of degrees vs. radians, I do have a calculator that computes the trig functions for angles in degrees. On the other hand, the computer language python has a math function which computes trig functions for angles in radians. (They do have a math function for pi, which provides a simple way of writing a function for the conversion.) I have always seen the special right triangles in geometry, 30°-60°-90° and 45°-45°-90°, given in degrees, not radians.
@@jimlocke9320 If you want to be coherent, x here has no unit, it is simply a positive real number, it is the ratio of the length of AE in a certain unit divided by this unit. this unit can be meters, millimeters, miles, ... That's the reason why you can draw the triangle as big or as small as you want on your paper.
Considering degrees, if you were a french student using degrees you would get a bad mark. But I know that in anglo-american countries it is still used and I also often use degrees here when I try to "speak" english (when there is no special problem with that unit).
@@marcgriselhubert3915 Does that mean that the special 30°-60°-90° right triangle is designated π/6 - π/3 - π/2 and the special 45°-45°-90° isosceles right triangle is designated π/4 - π/4 - π/2 in France?
@@jimlocke9320 The first "special triangle" has no name here and the second is called "rectangle isoscele".
SCOOP: In fact, this problem is physicaly impossible. A right triangle ABC with hypotenuse [A,C] divided by points E and D in 3 equal lengthes represented by the real x, with a = EB = sin(x) and b = DB = cos(x) doen't exist, it is not possible to construct it.
If this triangle was possible to construct, then we proved that necessary x = sqrt(5)/5, and then a = sin(x) = 0.4324 (about) and b = cos(x) = 0,9016 (about)
Then B is at the intersection of the circle of center E and of radius a with the circle of center D and of radius b.
a+b = 1.334, abs(a-b) = 0.4692 and DE = sqrt(5)/5 = 0.447. We do not have abs(a-b) < DE< a+b, so these two circles do not intersect, the construction is impossible. (It is also possible to verify that the circle of diameter [A,C], normaly containing B, does not intersect with the two precedent circles)
So there was really a problem with this exercice, but not with sin and cos of a real.
We can use the cosine rule for triangles. So let AB be h and the angle EAB be theta. Then cos(theta) = h/(3x). Applying the cosine rule for triangle AEB yields a^2 = x^2 + h^2 - (2xh)cos(theta) and for triangle ABD gives us b^2 = (2x)^2 + h^2 - (4xh)cos(theta). Substituting a = sin(x) and b = cos(x) and cos(theta) = h/(3x) gives us the two equations: sin(x)^2 = x^2 + h^2 - (2/3)h^2 and cos(x)^2 = 4(x^2) + h^2 - (4/3)h^2. Next we add these two equations giving us 1 = 5(x^2) yielding x = sqrt(5)/5.
We use an orthonormal center B and first axis (BC), we have B(0; 0) A(0; a') C(c; 0) E(c/3; (2/3).a') D((2/3).c; a'/3) with a' = BA and c = BC
BE^2 = (1/9).(c^2 +4.(a'^2)) = (sin(x))^2 and BD^2 = (1/9).(a'^2 +4.(c^2)) = (cos(x))^2. As (sin(x))^2 + (cos(x))^2 = 1, we have (1/9)(5.(a'^2) +5.(c^2)) = 1
Then a'^2 + c^2 =9/5. As a'^2 + c^2 = BA^2 + BC^2 = AC^2, we the have that AC^2 = 9/5 and AC = 3/sqrt(5). Finally x = AC/3 = 1/sqrt(5) = (sqrt(5))/5.
In ∆ABC, AB²+BC²=9X²
and a²+b²=1
Applying Stewart's Theorem to ∆ABD and ∆EBC
AB²X+b²X=2X(a²+X²)
BC²X+a²X=2X(b²+X²)
Adding them,
5X³=X
X=1/√5
These equations are not correct!
@@Emerson_Brasil Why?
@@harikatragadda
xAB² + b²x =2x(a²+x²)
e
xBC² + a²x =2x(b²+x²)
The correct equations!
@@Emerson_Brasil Corrected, thanks!
a^2+b^2=1 by definition. Therefore triangle EDB must be right triangle with hypotenuse 1. Therefore x=1. And angle EBD must be right angle, but that’s contradict to the conditions
When we have a triangle with sides a,b,c we can generate a formula for the length of a median. Expand that median to generate a parallelogram and we know that the sum of the squares of the sides equels the sum of the squares of the diagonals. The median BN in triangle ABC is 1.5x . And now, in triangle EBD apply the formula.
a^2 = 4m^2 + n^2; b^2 = m^2 + 4n^2 → m^2 + n^2 = 1/5 = x^2 → x = √5/5
btw: φ = 30° → BCA = θ → b > a↔ θ < 3φ/2
if θ = 3φ/2 → a = b → m = n; amazing: it works for any angle 🙂
As someone has pointed out, the answer does not satisfy the triangular inequality. In other words, I solved it using cosine rule and got 1/sqrt(5), but also found that cos(A) is not a real number, its square being equal to -0.0216
Bom dia Mestre
Essa foi difícil 😮
Grato pelos ensinamentos
Questão linda !
x=sqrt(0.2)=.4472 approx
a = sin x, so x= inv sin a, meaning a = 26.57 degrees or .4636 radians
b = cos x, so x= inv cos b, meaning b= 63.43 degrees or 1.1071 radians
a - b is greater than x, whether x is in radians or degrees, meaning the triangle EDB cannot be constructed.
Clearly a mistake in setting the question. The mistake is using x as both the angle for calculating a and b, and for the dimensions of AE, ED and DC. They are unrelated quantities.
My only question is whether or not x is in degrees or radians when we take sin(x) and cos(x).
x is a (positive) real. Sin(x) and cos(x) have a meaning for x real. For x real, sin(x) = sin(angle (x rad mod 2.Pi)), id for cos. Normally we should always use radians and never degrees or other angle units to avoid this type of question.
Let's find x:
.
..
...
....
.....
First of all we add the points F and G on BC such that BEF and BDG are right triangles. Since ABC is also a right triangle, we can apply the Pythagorean theorem to all these triangles:
AC² = AB² + BC²
BE² = EF² + BF²
BD² = DG² + BG²
(3x)² = AB² + BC²
sin²(x) = (2*AB/3)² + (BC/3)²
cos²(x) = (AB/3)² + (2*BC/3)²
9x² = AB² + BC²
sin²(x) = 4*AB²/9 + BC²/9
cos²(x) = AB²/9 + 4*BC²/9
9x² = AB² + BC²
sin²(x) + cos²(x) = 4*AB²/9 + BC²/9 + AB²/9 + 4*BC²/9
1 = 5*AB²/9 + 5*BC²/9 = 5*(AB² + BC²)/9 = 5*(9x²)/9 = 5*x² ⇒ x = 1/√5
Best regards from Germany
Hi, Do you have any idea of the value of the smallest angle ACB? (my first guess is: 18 degrees or pi/10 rad)
But now I don't think it's possible to calculate the angle!
I got n² + m² = 0.2 and n² - m² = cos(2x)/3 = 0.208655 and therefore m² = -0.1043
O segmento a é perpendicular ao segmento AD, visto que divide AD em duas partes iguais, logo, AB = a. Então temos dois triângulos retângulos. Por Pitágoras, igualamos as duas equações e chegamos ao resultado posto pelo professor. Linda questão.
@ 2:15 Onomatopoetic. "Achoo!", I need Sinx! 🤧 ...it's the weekend and I act like it! Let's party! Moooooooooo 🙂
X=√5/5 units.
1 / 3 root 2 ?
There are many solutions: x = √5/5 + 2kπ, k = 0,1,2, ...
This is not a very good problem though.
No, there is only one solution: (sqrt(5))/5, as obtained in the proof.
Imagine k = 1,000,000 for example. The ABC triangle will be enormous but BE and BD always between 0 and 1... !
a=sinx
Sinx is meaningless as x is not any angle.
Sin(x) and cos(x) have a meaning for x real.
@@marcgriselhubert3915 So what are the lengths of a & b?
@@montynorth3009 a = sin((sqrt(5))/5) which is about 0.432, and b = cos((sqrt(5))/5) which is about 0.902, naturaly.