CDAº=140º-θ→ ADBº=40º+θ→ BADº=60º-θ → Trazamos la paralela a BC por A y el simétrico de AC según el eje vertical que pasa por A→ Angulo llano en A =180º=θ+40º+60º-θ+40º+θ =θ+140º→ θ=180º-140º=40º. Gracias y un saludo.
1. Drop a perpendicular from A on BC ( point E) and extend it to intersect a line From B under 80 degrees angle(point F) so that BF=AB, AF=EF. 2. Triangles AED and FED are equal. BD is perpendicular to AF as diagonals , therfore ABFD is a rhombus. Therefore, AB=FD=BF=AD =DC. Triangle ADC is equilateral, angle CAD=40degrees = angle DCA.
I don't see that. Dropping a line from point A to point F that is perpendicular to BC and angle EBF = 80 deg will ensure that: 1. AB = BF 2. AD = DF 3. Triangles AED and FED are equal 4. BD is perpendicular to AF However, it does not follow that ABFD is a rhombus nor that AB = AD. Point D can move along the horizontal axis between E and C and preserve 1 - 4 without forming a rhombus. It does work out in this example that ABFD is a rhombus, but it cannot be assumed on the basis of showing that the intersecting diagonals are perpendicular. Any quadrilateral where the two adjacent sides are equal will result in perpendicular diagonals, e.g., a "kite".
@@rms3 i tried to prove the rhombus-unsuccessfully. Then i tried another method: bisect the 80 degrees angle. Point K on AC. Consider two triangles: BKC and ADC.They are congruent.Therefore AC=BC, then theta is 20 degrees. Am I right?
@@ludmilaivanova1603 OK, so if: 1. Angle BKC = DAC = 40 deg 2. Both triangles share angle ACD = KCB = theta 3. The obtuse angle in both triangles ACD and CKB = 140 - theta 4. So, the two triangles are similar 5. The side opposite 40 deg in triangle ACD (CD) = KC, by Law of Sines 6. Therefore, the two triangles are congruent 7. Since KC = KB, BKC is an isoceles triangle, so theta = angle KBC = 40 deg Yes, that seems to work.
@@ludmilaivanova1603 OK, on further thought, just proving the that triangles ADC and BKC are congruent does not prove that they are isoceles. So, I guess your method does not work after all. The solution from comment below seems to work: "Using the law of sines in triangle ABD we find AB/sin(θ+40)=AD/sin80 and using the law of sines in triangle ADC we find AD/sin(θ)=AB/sin40 and from this sin(θ+40)*sin( θ)=sin(40)*sin(80) and from this θ=40"
Using the law of sines in triangle ABD we find AB/sin(θ+40)=AD/sin80 and using the law of sines in triangle ADC we find AD/sin(θ)=AB/sin40 and from this sin(θ+40)*sin( θ)=sin(40)*sin(80) and from this θ=40
You are right. It is impossible to draw in a pure geometric way an angle equal to 40°+theta on AD. The second method appears only when you obtain the solution through method 1
The angle theta is 40°. Believe it or not I actually understood the first method better than the second method. I guess that it was because I considered that a refresher of a fascinating trig identity. And not only that but I think that it was the pair of vertical angles that made the two methods similar to each other if I understood correctly. And the congruence postulate was AAA all thanks to the exterior angle theorem as well as the vertical angles. Two angles are congruent and equal the same angle while the other angles are congruent due to vertical angles and Law of Sines right???
EC construction in 2 stages, first draw angle 40+theta at arbitrary point on AD extension and draw a parallel from C to meet the extension at E.
CDAº=140º-θ→ ADBº=40º+θ→ BADº=60º-θ → Trazamos la paralela a BC por A y el simétrico de AC según el eje vertical que pasa por A→ Angulo llano en A =180º=θ+40º+60º-θ+40º+θ =θ+140º→ θ=180º-140º=40º.
Gracias y un saludo.
Awesome.Thank you professor!
Sine rule:
sin θ / AD = sin 40°/ x
sin 80° /AD = sin (θ+40°) /x
Dividing:
sinθ/sin80°=sin40°/sin(θ+40°)
θ = 40° ( Solved √ )
θ + (θ+40°) = 40° + 80°
2θ = 80°
θ = 40°
1. Drop a perpendicular from A on BC ( point E) and extend it to intersect a line From B under 80 degrees angle(point F) so that BF=AB, AF=EF.
2. Triangles AED and FED are equal. BD is perpendicular to AF as diagonals , therfore ABFD is a rhombus. Therefore, AB=FD=BF=AD =DC.
Triangle ADC is equilateral, angle CAD=40degrees = angle DCA.
I don't see that. Dropping a line from point A to point F that is perpendicular to BC and angle EBF = 80 deg will ensure that:
1. AB = BF
2. AD = DF
3. Triangles AED and FED are equal
4. BD is perpendicular to AF
However, it does not follow that ABFD is a rhombus nor that AB = AD. Point D can move along the horizontal axis between E and C and preserve 1 - 4 without forming a rhombus.
It does work out in this example that ABFD is a rhombus, but it cannot be assumed on the basis of showing that the intersecting diagonals are perpendicular. Any quadrilateral where the two adjacent sides are equal will result in perpendicular diagonals, e.g., a "kite".
yes, you are right. It is not enough evidence that this is a rhombus. It has to be proved.
@@rms3 i tried to prove the rhombus-unsuccessfully. Then i tried another method: bisect the 80 degrees angle. Point K on AC. Consider two triangles: BKC and ADC.They are congruent.Therefore AC=BC, then theta is 20 degrees. Am I right?
@@ludmilaivanova1603 OK, so if:
1. Angle BKC = DAC = 40 deg
2. Both triangles share angle ACD = KCB = theta
3. The obtuse angle in both triangles ACD and CKB = 140 - theta
4. So, the two triangles are similar
5. The side opposite 40 deg in triangle ACD (CD) = KC, by Law of Sines
6. Therefore, the two triangles are congruent
7. Since KC = KB, BKC is an isoceles triangle, so theta = angle KBC = 40 deg
Yes, that seems to work.
@@ludmilaivanova1603 OK, on further thought, just proving the that triangles ADC and BKC are congruent does not prove that they are isoceles. So, I guess your method does not work after all.
The solution from comment below seems to work:
"Using the law of sines in triangle ABD we find AB/sin(θ+40)=AD/sin80 and using the law of sines in triangle ADC we find AD/sin(θ)=AB/sin40 and from this sin(θ+40)*sin( θ)=sin(40)*sin(80) and from this θ=40"
Using the law of sines in triangle ABD we find AB/sin(θ+40)=AD/sin80 and using the law of sines in triangle ADC we find AD/sin(θ)=AB/sin40 and from this sin(θ+40)*sin( θ)=sin(40)*sin(80) and from this θ=40
I did it as your method 1 by my goto method. But your method 2 is very ingenius.
You are right. It is impossible to draw in a pure geometric way an angle equal to 40°+theta on AD. The second method appears only when you obtain the solution through method 1
cos(120°)=cos(40°+2θ)→θ=40°+kπ (k is a natural number), and since θ
@ 9: 28 I have doubts regarding construction of an unknown angle, not sure its leg will hit the point C.
{80°B+40°A}=120°BA {120°BA+60°C}=:180°BCA 3^60 3^20^2 3^2^10^2 3^1^2^5^1 3^2^5^1 3^2^1^1 3^2 (BCA ➖ 3BCA+2).
The angle theta is 40°. Believe it or not I actually understood the first method better than the second method. I guess that it was because I considered that a refresher of a fascinating trig identity. And not only that but I think that it was the pair of vertical angles that made the two methods similar to each other if I understood correctly. And the congruence postulate was AAA all thanks to the exterior angle theorem as well as the vertical angles. Two angles are congruent and equal the same angle while the other angles are congruent due to vertical angles and Law of Sines right???
Fiz contruindo um paralelogramo. Depois, arco capaz e enfim por ângulo inscrito.
Thanks 🙏👍💯😊
Delta=40°
40
Law of sines
x/sin(40) = AD/sin(t)
x/sin(40+t) = AD/sin(80)
AD/x = sin(t)/sin(40)
AD/x = sin(80)/sin(40+t)
sin(t)/sin(40) - sin(80)/sin(40+t) = 0
(sin(t)sin(40+t) - sin(80)sin(40))/(sin(40)sin(40+t)) = 0
sin(t)sin(40+t) - sin(80)sin(40) = 0
sin(t)(sin(40)cos(t)+cos(40)sin(t)) - sin(80)sin(40) = 0
sin(40) sin(t)cos(t) +cos(40)sin^2(t) - sin(80)sin(40) = 0
sin(40) tan(t) +cos(40)tan^2(t)- sin(80)sin(40)(1+tan^2(t)) = 0
(cos(40) - sin(80)sin(40))tan^2(t) - sin(40) tan(t) - sin(80)sin(40)=0
This gives us two results but one must be rejected
theta = 40
It appears that ADC is isosceles
4:30 Is it possible to say 40°+θ=80° θ=40° from the arguments of the sin()-function?
Yes, you can directly compare there, if you need single solution.
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