It's interesting that if a=b (contrary to the stated condition), you have a parallelogram with angles 30 deg, 150 deg, 30 deg, 150 deg. a and b can now be any positive amount, so a+b is undetermined.
Thanks for first method. Very insightful. Second method was the one I would have gone trough just algebra autopilot but the first one with flipping the triangle to get collinear points was lovely.
@JoanRosSendra, after we have ∆APQ with AP=PQ=5, AQ=a+b and angle APQ = 180º - 30º - 30º = 120º, then by cosine rule we obtain, AQ² = AP² + PQ² - 2.AP.PQ. cos 120º; (a + b)² = 25+25 - 2.5.5(-½) =75, and so on
Cortamos la figura por PC y alineamos a con b, quedando P y C superpuestos y la nueva figura es un triángulo isósceles BCA---> BA=a+b =2*[5√3/2] =5√3. Gracias y saludos
Reflect BPC along side PC. The quadrilateral AB’PC is a regular trapezoid inscribed in a circle and the diagonals AP and B’C are both equal two 5. Quite easy to demonstrate with similar triangles that a+b = 5 √ 3… and PC^2 = 25 - a.b
Connect P and C. If we place the left triangle on top of the right triangle the way the sides=5 coincide and sides a and b coincide due to angle 30 degrees, then consider "a" and "b" difference (b-a) as the base of an isosceles triangle( PC is the same for both triangles), drop a perpendicular from point P , it is 2.5, and use Pythagorean theorem ((b-a)/2 +a)^2 ) + 2,5^2=5^2 (a+b)^2=75 ; a+b = 5sqrt3
I reflect triangles APC and BPC along sides AC and BC respectively. This resulted in Equilateral triangles APP' and PBP''. From this I used the Pythagorean theorem with two triangles of equal hypotenuse to find (a + b) = 5sqrt(3).
Prolongamos el segmento AC hasta un punto Q desde el cual trazamos un ángulo de 30 grados hasta el punto P. Habremos formado un triángulo isosceles AQP. Por tanto, PQ=PA=5 Trazamos el segmento PC y tendremos dos triángulos congruentes (PBC y CQP) ya que comparten el ángulo de 30 grados, el lado opuesto y un lado de 5. Por tanto, CQ=BP=a. Así pues, AQ=a+b, que es nuestro objetivo... Trazamos la altura desde P a AQ en el triángulo isosceles APQ que nos dividirá AQ en dos mitades, es decir (a+b)÷2 El coseno de 30 es igual a [(a+b)÷2]÷5=(raíz de 3)÷2. ===》a+b=5 raíz de 3
Thank you for the problem and its solution. An alternative solution would be: As the [angle PAC] = [angle PBC] and |BC| = |AP|, one might try to make use of congruency, which can e.g. be done by choosing a point "Q" inside the segment AC s.t. |AQ| = |BP|, which means that triangles PBC and QAP are congruent, and that |PC| = |PQ|, which means that triangle CPQ is isosceles. If one connects P with the midpoint of CQ, let's call it "M", then, due to [angle PAC] = 30°, |AM| = 5*sqrt(3)/2. Now, |AQ| = a due to the congruency of the triangles PBC and QAP, |AC| = b by the conditions of the problem, so what is |AM| equal to? Well, as M is the midpoint of QC, it is equal to the arithmetic mean of a and b. Thus (a+b)/2 = |AM| = 5*sqrt(3)/2. As a result, a+b = 5*sqrt(3).
This start was not needed: Drawing PE at right-angles to AC gives PE = ½AP in the right-angled triangle PEA After constructing a point Q so that PQ is parallel to CA and PQ = a, here is a trapezium AQPC. [AQPC] is ½(a+b) × PE and PE is 2.5 a+b = [AQPC] / 1.25 = 0.8×[AQPC] Because PQ is parallel to CA there is a 30º angle at P which is alternate with angle PAC . This makes triangle PAQ congruent with triangle BCP [AQ PC] = [ACP] + [BCP] Using the area formula with the original two 30º angles given shows that these areas are (5/4)a and (5/4)b which is already known. Then I found method two and got almost there with one mistake to correct Now PC is common to these triangles so its square can be written twice in terms of cos 30º . PC^2 = a^2 +5^2 -5a(3^½) = b^2 + 5^2 - 5b (3^½) a^2 + 25 - 5a (3^½)/2 = b^2 +25 - 5b (3^½) from here b^2- a^2 = (b+a)(b-a) = 5 (b-a)(3^½) b+a = 5(3^½) b = 5(3^½) a+b = 5 × square root of 3
Someone please explain this riddle to me! To this day, when this guy posts the solution, I still don't understand it. Is it math from Jupiter or Mars?😂😂😂
c = Segment PC
Cosine rule, twice :
c² = 5²+a²-2*5*a*cos30°
c² = 5²+b²-2*5*b*cos30°
Equalling :
a² - 5√3 a = b² - 5√3 b
a² - b² = 5√3 (a-b)
(a+b)(a-b)/(a-b)= 5√3
a + b = 5√3 cm ( Solved √ )
Yo lo hice igual!!! Too easy
It's interesting that if a=b (contrary to the stated condition), you have a parallelogram with angles 30 deg, 150 deg, 30 deg, 150 deg. a and b can now be any positive amount, so a+b is undetermined.
Thanks for first method. Very insightful.
Second method was the one I would have gone trough just algebra autopilot but the first one with flipping the triangle to get collinear points was lovely.
@JoanRosSendra, after we have ∆APQ with AP=PQ=5, AQ=a+b and angle APQ = 180º - 30º - 30º = 120º, then by cosine rule we obtain, AQ² = AP² + PQ² - 2.AP.PQ. cos 120º;
(a + b)² = 25+25 - 2.5.5(-½) =75, and so on
Cortamos la figura por PC y alineamos a con b, quedando P y C superpuestos y la nueva figura es un triángulo isósceles BCA---> BA=a+b =2*[5√3/2] =5√3.
Gracias y saludos
Reflect BPC along side PC. The quadrilateral AB’PC is a regular trapezoid inscribed in a circle and the diagonals AP and B’C are both equal two 5.
Quite easy to demonstrate with similar triangles that a+b = 5 √ 3… and PC^2 = 25 - a.b
Connect P and C.
If we place the left triangle on top of the right triangle the way the sides=5 coincide and sides a and b coincide due to angle 30 degrees, then consider "a" and "b" difference (b-a) as the base of an isosceles triangle( PC is the same for both triangles), drop a perpendicular from point P , it is 2.5, and use Pythagorean theorem
((b-a)/2 +a)^2 ) + 2,5^2=5^2
(a+b)^2=75 ; a+b = 5sqrt3
I reflect triangles APC and BPC along sides AC and BC respectively.
This resulted in Equilateral triangles APP' and PBP''.
From this I used the Pythagorean theorem with two triangles of equal hypotenuse to find (a + b) = 5sqrt(3).
Prolongamos el segmento AC hasta un punto Q desde el cual trazamos un ángulo de 30 grados hasta el punto P.
Habremos formado un triángulo isosceles AQP. Por tanto, PQ=PA=5
Trazamos el segmento PC y tendremos dos triángulos congruentes (PBC y CQP) ya que comparten el ángulo de 30 grados, el lado opuesto y un lado de 5. Por tanto, CQ=BP=a. Así pues, AQ=a+b, que es nuestro objetivo...
Trazamos la altura desde P a AQ en el triángulo isosceles APQ que nos dividirá AQ en dos mitades, es decir (a+b)÷2
El coseno de 30 es igual a [(a+b)÷2]÷5=(raíz de 3)÷2. ===》a+b=5 raíz de 3
Thank you for the problem and its solution. An alternative solution would be: As the [angle PAC] = [angle PBC] and |BC| = |AP|, one might try to make use of congruency, which can e.g. be done by choosing a point "Q" inside the segment AC s.t. |AQ| = |BP|, which means that triangles PBC and QAP are congruent, and that |PC| = |PQ|, which means that triangle CPQ is isosceles. If one connects P with the midpoint of CQ, let's call it "M", then, due to [angle PAC] = 30°, |AM| = 5*sqrt(3)/2. Now, |AQ| = a due to the congruency of the triangles PBC and QAP, |AC| = b by the conditions of the problem, so what is |AM| equal to? Well, as M is the midpoint of QC, it is equal to the arithmetic mean of a and b. Thus (a+b)/2 = |AM| = 5*sqrt(3)/2. As a result, a+b = 5*sqrt(3).
Clearly and thoroughly explained for both methods, Thank you.
This start was not needed:
Drawing PE at right-angles to AC gives PE = ½AP in the right-angled triangle PEA
After constructing a point Q so that PQ is parallel to CA and PQ = a, here is a trapezium AQPC. [AQPC] is ½(a+b) × PE and PE is 2.5
a+b = [AQPC] / 1.25 = 0.8×[AQPC]
Because PQ is parallel to CA there is a 30º angle at P which is alternate with angle PAC . This makes triangle PAQ congruent with triangle BCP
[AQ PC] = [ACP] + [BCP] Using the area formula with the original two 30º angles given shows that these areas are (5/4)a and (5/4)b which is already known.
Then I found method two and got almost there with one mistake to correct
Now PC is common to these triangles so its square can be written twice in terms of cos 30º .
PC^2 = a^2 +5^2 -5a(3^½) = b^2 + 5^2 - 5b (3^½) a^2 + 25 - 5a (3^½)/2 = b^2 +25 - 5b (3^½)
from here
b^2- a^2 = (b+a)(b-a) = 5 (b-a)(3^½) b+a = 5(3^½) b = 5(3^½)
a+b = 5 × square root of 3
2nd solution is very good.Thank you professor.
Col semplice teorema del coseno risulta a^2+5^2-10acos30=b^2+5^2-10bcos30...divido per (a-b),(quindi a diverso da b)...a+b=10cos30=5√3
Sir,
Like new solu:
PAC🔺,PBC🔺
pc=pc
B°=A°
ත්රිකෝණ දෙක එක මත එක තැබීමෙන්,
PA මතට BC වන පරිදි,
PD ලබ්භකය,CA මතට අදින්න,
AC=b,
BP=a,
PC=(b-a)/2
5.cos30°=(b-a)
5root3/2=a+b/2
5root3=a+b
Why is alpha plus beta equal to 180 degree in 1.st method?
Because sine is cyclic, so sin(x) = sin(180°-x)
Because sin(β) = sin(180−β)
Therefore, if sin(α) = sin (β) = sin(180−β), then either
α = β
α = 180−β → α+β = 180
Thanks for your replies guys. Actually, I was able to get the point by myself after writting my comment. Lots of thanks anyway.
Second method is much better
(5)^2=25 {30°A30°B+90°C}=150°ABC 150°ABC/25=6 (ABC ➖ 6ABC+6).
Someone please explain this riddle to me! To this day, when this guy posts the solution, I still don't understand it. Is it math from Jupiter or Mars?😂😂😂
BC is √3 *(a+b)
pl explain how Alpha and beta is 180deg
Because sin(β) = sin(180−β)
Therefore, if sin(α) = sin (β) = sin(180−β), then either
α = β
α = 180−β → α+β = 180
Sorry, a+b = √3 * BC
Method 2 is far superior to method 1.
péssima tradução! não dá para entender.
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Kya aap ka name satyam hai