Let O = center of circle. OA = OP = OQ = OB = radius = x/2 Let ∠AOP = ∠POQ = θ → ∠BOQ = 180 − 2θ Using law of cosines in △AOP we get: cos θ = [(x/2)²+(x/2)²−3²]/[2(x/2)(x/2)] = (x²−18)/x² Using law of cosines in △BOQ we get: cos(180−2θ) = [(x/2)²+(x/2)²−7²]/[2(x/2)(x/2)] = (x²−98)/x² Now we setup relationship between these angles to get equation in x only cos 2θ = −cos(180−2θ) 2 cos²θ − 1 = −(x²−98)/x² 2 (x²−18)²/x^4 − 1 = (98−x²)/x² 2 (x²−18)² − x^4 = x² (98−x²) 2 (x²−18)² − x^4 = 98x² − x^4 2 (x²−18)² = 98x² (x²−18)² = 49x² (x²−18)² − 49x² = 0 (x²−7x−18) (x²+7x−18) = 0 (x+2) (x−9) (x−2) (x+9) = 0 x = ±2, x = ±9 Since x is diameter of circle, it must be longer than chords → x > 7 *x = 9*
@ 6:53 connect A and Q and get a right triangle AQM (angle ACB=90 degrees) which is similar to BPM. from that, 6/a= 7+a/3. 18=7a+a^2. a=2. sinse triangle ABC is isosceles D= BM = 9.
The answer is 9 units. And it looks like this a good example of knowing what would be the right circle theorem to make use of in order to justify auxiliary lines. I am kind of wondering, how many circle theorems justify auxiliary lines???
A fast solution can be drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x. We can find x with Euclid’s theorem on APB right triangle: 3^2=x*(x-7)/2 x^2-7x-18=0 X=9 Very nice problem, very nice the solution by Mathboost❤
soli9mana, would you describe in greater (clearer) detail your construction? I dont' understand the "drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x" (1) I know what an 'isoscele trapezoid' is; I don't understand HOW you got it. (2) I tried to look up "Euclid's theorem", but can only find a theorem from number theory, not one that would apply to right triangles... HELP!! Thx. 😊
@ Sure. You only have to draw a little different figure in which the segments whose length is 3 are one from A and the other from B, at the opposite sides of the diameter, simmetrically. Then the segment of 7 on the top, in the middle, parallel to the diameter, being lateral sides at the same height. In this way you see an isosceles trapezoid , with the diameter, and your goal is to find its major base. I hope it’s clearer, can’t say better
@@timeonly1401 1) I simply built the isosceles trapezoid with the recombined segments (chords if you prefer) of the original figure: instead of having the sequence 3-3-7, replace with 3-7-3. Or try to draw an isosceles trapezoid inscribed in the semicircle with the major base x, minor base 7 and the two sides 3, in this way you will have your modified figure. 2) If you are a fan of geometry problems, you should know the two beautiful theorems of Euclid, the man who wrote the most beautiful mathematics book in the world! Know that there are a couple that concern right triangles, the one I used in the problem exposed says that: In any right triangle, the area of the square on a side adjacent to the right angle is equal to the area of the rectangle whose dimensions are the length of the projection of this side on the hypotenuse and the length of the hypotenuse. I Hope this help
Knowing cos(theta) = 7/(2r) and then using the Law of Cosines with one of the triangles with two sides "r" and third side "3" one can determine that 2r is 9. Theta is the angle subtended by the "3" side and the center of the semicircle.
Adding -{lines AQ and}- BP to the diagram and labelling O the centre of AB ( X is the diameter of the semicircle and O is the centre.) , -{ then APB and AQB are two rightangled triangles with 90º angles at P and at Q respectively. AQ^2= X^2 - 7^2 and BP^2 = X^2 - 3^2 by Pythagoras' theorem. AC.CQ is equal to BC.CP if we label the point of intersection of AQ and BP as C. A, P, Q and B are given points on a (semi-)circle. (AC+CQ)^2 = AQ^2 = AC^2 +CQ^2 +2AC.CQ (BC+CP)^2 = BP^2 = BC^2 + PC^2 + 2BC.CP Now BP^2 - AQ^2 is equal to BC^2 +PC^2 - AC^2 - CQ^2 +(2BC.CP- 2AC.CQ , this term in the bracket equals zero)}- but I think there should be a solution that does not use AQ and does not use C Let ½X = r the radius, which equals OA, OP, OQ and OB Now angles AOP and POQ are equal and naming them each tº then the value of angle QOB is 180º- 2tº and 7x7 = 49 = r^2 +r^2 - 2.r.r cos(180-2t) and 3x3 =9 = r^2 +r^2 - 2.r.r cos (t) 49-9 = 2.r.r ( cos (t) - cos (180 -2t) r.r = r^2= (40/2) / ( cos(t) + cos (2t)) cos (2t) = cos^2(t)-sin^2(t) = 2cos^2(t ) -1 r^2 = 20 / (2cos^2(t) +cos(t) -1) the denominator factorises as ( 2c+1)(c-1) where c= cos(t) So a couple of paths i have shown here which are to be abandonned and the solution to be sought in the video provided here. Peeping at it now I perceive that I forgot that construction can be extended beyond the given shape, and this can simplify the solution. Thank you again, Math Booster , well done to students who " got it" on their own efforts, and encouragement and better "luck" next time to others!
Draw segments AQ & PB, and call their lengths c & d, respectively. ∠APB & ∠AQB are right angles (By Thales Thm). So, applying Pythagorean Thm to ∆AQB & ∆APB: c² + 49 = x² and 9 + d² = x², so that c=√(x²-49) and d=√(x²-9). All the vertices of quadrilateral APQB are on the semicircle (&, hence, the circle), so APQB is a CYCLIC quadrilateral. So, the product of the diagonals is equal to the sum of the products of opposite sides (Ptolemy's Thm). Applying, we get: (AQ)(PB) = (AP)(QB) + (PQ)(AB) c d = (3)(7) + (3)x = 3(x+7) Subbing in c and d from earlier: √(x²-49) √(x²-9) = 3(x+7) Squaring both sides and solving for x: (x²-49)(x²-9) = 9(x+7)² (x-7)(x+7)(x²-9) = 9(x+7)² [x is a diameter which is the longest chord in the circle, so x > 7. Hence x-7 > 0. And because x-7 is not zero, we can divide both sides by it.] (x-7)(x²-9) = 9(x+7) x³ - 7x² - 9x + 63 = 9x + 63 x³ - 7x² - 18x = 0 x(x² - 7x -18) = 0 [x > 7 ⇒ x ≠ 0] x² - 7x -18 = 0 (x-9)(x+2) = 0 x = 9 or x = -2 (elim., since x>7) so x=9. Done!!
You can simplify this problem by rearranging the line segments with the 7 long one at the center. Then symmetry gives 2 triangles: one hypotenuse 3, one leg r-3.5, & the other hypotenuse r, one leg 3.5. Equating Pythagoras for the 2 other legs gives equation 3² - (r - 3.5)² = r² - (3.5)². 2r² - 7r - 9 = 0. r = 4.5. X = 2r = 9.
@@marcusdecarvalho1354 Re-arranging the order of the line segments does not affect the radius, so they can be re-arranged to exploit symmetry to simplify the problem.
@@kateknowles8055 For example, in the third line of working, by the double angle formula for sine sin(2 arcsin(3/x)) = 2 sin(arcsin(3/x)) cos(arcsin(3/x)) The sin and arcsin cancel out so the first term on the right is just 3/x. The second term is cos of something that 3/x is the sin of, so using cos^2 = 1 - sin^2 you get sqrt(1 - 9/x^2). With sin(pi/2 - arcsin(7/x)), you have the complementary angle so it is cos(arcsin(7/x)) and proceed similarly.
We use an orthonormal center O, the center of the circle, and first axis (OA). We have A(R; 0) B(-R; 0) P(R.cos(t); R.sin(t)) Q(R.cos(2.t); R.sin(2.t)) with R the radius of the circle and t = angleOAP, 0°
Or, instead...... 1) Connect AQ. 2) Drop perpendicular from P to AQ, intersecting AQ at R. 3) Extend PR through O to intersect circle at S, making PS a diameter (length X), with PR = a, and RS = X - a. 4) For Right△APR, calculate (by Pythagoras) the length of AR as √(3² - a²) = √(9 - a²). 5) Use Intersecting Chords (AQ and PS) to figure 'a' in terms of X... i.e. AR * RQ = PR * RS: √(9 - a²) * √(9 - a²) = a * (X - a), or 9 - a² = aX - a², so that 9/X = a 6) Next, use Right △AQB and Pythagoras to establish that: 7² + [2 * √(9 - a²)]² = X², which is, 49 + 36 - 4a² = X²; simplified to 'a' in terms of X becomes: √[(85 - X²)/4] = a 7) Substituting for 'a' from #5, we get: √[(85 - X²)/4] = (9/X), square both sides to get (85 - X²)/4 = 81/X², or 85X² - X⁴ = 324, or X⁴ - 85X² + 324 = 0 8) Substituting Z = X² we have: Z² - 85Z + 324. 9) Use the quadratic equation to solve for Z: Z = [85 ± √(85² - 4 * 1 * 324)]/2, or Z = [85 ± √(7225 - 1296)]/2 Z = [85 ± √(5929)]/2 = [85 ± 77]/2, or Z = 81 or 4 10) Reverting with Z = X² we get: X = √81 = 9, or X = √4 = 2 Obviously, X cannot be equal to 2, as the resulting △AQB would be degenerate. Therefore, X = 9. Q.E.D.
Hay algo que no me cuadra. El problema planteado tiene infinitas soluciones si se resuelve con un compás y una regla, dependiendo del ángulo que arbitrariamente formemos entre los segmentos PA y PQ y, por ende, de la longitud del segmento AQ. Cuando un problema geométrico tiene una 'única' solución trigonométrica, tiene tambíen una 'única' solución gráfica. Y no es el caso.
🎉🎉🎉 A questão é muito boa. Eu resolvi aplicando o Teorema dos Cossenos e a minha solução não exige uma construção auxiliar. Parabéns pela escolha 🎉🎉🎉 BRASIL Novembro de 2024.
APQ=α...teorema del coseno 18-18cosα=x^2+49+14xcosα..x^2+31+(14x+18)cosα=0...poi,teorema del seno √(x^2-49)/sinα=3/sin(90-α/2)=3/cosα/2..(x^2-49)/(1-(cosα)^2)=9/(1+cosα)/2….semplifico risulta cosα=(67-x^2)/18..sostituisco mi risulta una cubica 7x^3-469x-882=0,con soluzione x=9
Applying Ptolemys Theorem we have 3•x+3•7=√(x^2-7^2)•√(x^2-3^2) so [3•(x+7)]^2=(x+7)•(x-7)•(x^2-9) , divide Both Sides by x+7>0, hence 9•(x+7)=(x-7)(x^-9) so 9x+63= x^3-9x-7x^2+63 so x^3-7x^2-18x=0,x>0 so x^2-7x-18=0 so (x+2)•(x-9)=0 so x=-2,x>0 Rejected ,or x=9 Accepted.
Let O = center of circle. OA = OP = OQ = OB = radius = x/2
Let ∠AOP = ∠POQ = θ → ∠BOQ = 180 − 2θ
Using law of cosines in △AOP we get:
cos θ = [(x/2)²+(x/2)²−3²]/[2(x/2)(x/2)] = (x²−18)/x²
Using law of cosines in △BOQ we get:
cos(180−2θ) = [(x/2)²+(x/2)²−7²]/[2(x/2)(x/2)] = (x²−98)/x²
Now we setup relationship between these angles to get equation in x only
cos 2θ = −cos(180−2θ)
2 cos²θ − 1 = −(x²−98)/x²
2 (x²−18)²/x^4 − 1 = (98−x²)/x²
2 (x²−18)² − x^4 = x² (98−x²)
2 (x²−18)² − x^4 = 98x² − x^4
2 (x²−18)² = 98x²
(x²−18)² = 49x²
(x²−18)² − 49x² = 0
(x²−7x−18) (x²+7x−18) = 0
(x+2) (x−9) (x−2) (x+9) = 0
x = ±2, x = ±9
Since x is diameter of circle, it must be longer than chords → x > 7
*x = 9*
@ 6:53 connect A and Q and get a right triangle AQM (angle ACB=90 degrees) which is similar to BPM. from that, 6/a= 7+a/3. 18=7a+a^2. a=2. sinse triangle ABC is isosceles D= BM = 9.
The answer is 9 units. And it looks like this a good example of knowing what would be the right circle theorem to make use of in order to justify auxiliary lines. I am kind of wondering, how many circle theorems justify auxiliary lines???
A fast solution can be drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x. We can find x with Euclid’s theorem on APB right triangle:
3^2=x*(x-7)/2
x^2-7x-18=0
X=9
Very nice problem, very nice the solution by Mathboost❤
soli9mana, would you describe in greater (clearer) detail your construction? I dont' understand the "drawing segment PQ on the right side simmetrically to AP and BQ in the middle getting an isosceles trapezoid with minor base 7 and lateral sides 3, major base x"
(1) I know what an 'isoscele trapezoid' is; I don't understand HOW you got it.
(2) I tried to look up "Euclid's theorem", but can only find a theorem from number theory, not one that would apply to right triangles...
HELP!! Thx. 😊
@ Sure. You only have to draw a little different figure in which the segments whose length is 3 are one from A and the other from B, at the opposite sides of the diameter, simmetrically. Then the segment of 7 on the top, in the middle, parallel to the diameter, being lateral sides at the same height. In this way you see an isosceles trapezoid , with the diameter, and your goal is to find its major base. I hope it’s clearer, can’t say better
@@timeonly1401 1) I simply built the isosceles trapezoid with the recombined segments (chords if you prefer) of the original figure: instead of having the sequence 3-3-7, replace with 3-7-3. Or try to draw an isosceles trapezoid inscribed in the semicircle with the major base x, minor base 7 and the two sides 3, in this way you will have your modified figure.
2) If you are a fan of geometry problems, you should know the two beautiful theorems of Euclid, the man who wrote the most beautiful mathematics book in the world! Know that there are a couple that concern right triangles, the one I used in the problem exposed says that:
In any right triangle, the area of the square on a side adjacent to the right angle is equal to the area of the rectangle whose dimensions are the length of the projection of this side on the hypotenuse and the length of the hypotenuse. I Hope this help
Knowing cos(theta) = 7/(2r) and then using the Law of Cosines with one of the triangles with two sides "r" and third side "3" one can determine that 2r is 9.
Theta is the angle subtended by the "3" side and the center of the semicircle.
Adding -{lines AQ and}- BP to the diagram and labelling O the centre of AB ( X is the diameter of the semicircle and O is the centre.) ,
-{ then APB and AQB are two rightangled triangles with 90º angles at P and at Q respectively. AQ^2= X^2 - 7^2 and BP^2 = X^2 - 3^2 by Pythagoras' theorem.
AC.CQ is equal to BC.CP if we label the point of intersection of AQ and BP as C. A, P, Q and B are given points on a (semi-)circle.
(AC+CQ)^2 = AQ^2 = AC^2 +CQ^2 +2AC.CQ
(BC+CP)^2 = BP^2 = BC^2 + PC^2 + 2BC.CP Now BP^2 - AQ^2 is equal to BC^2 +PC^2 - AC^2 - CQ^2 +(2BC.CP- 2AC.CQ , this term in the bracket equals zero)}-
but I think there should be a solution that does not use AQ and does not use C
Let ½X = r the radius, which equals OA, OP, OQ and OB Now angles AOP and POQ are equal and naming them each tº then the value of angle QOB is 180º- 2tº
and 7x7 = 49 = r^2 +r^2 - 2.r.r cos(180-2t) and 3x3 =9 = r^2 +r^2 - 2.r.r cos (t)
49-9 = 2.r.r ( cos (t) - cos (180 -2t) r.r = r^2= (40/2) / ( cos(t) + cos (2t)) cos (2t) = cos^2(t)-sin^2(t) = 2cos^2(t ) -1
r^2 = 20 / (2cos^2(t) +cos(t) -1) the denominator factorises as ( 2c+1)(c-1) where c= cos(t)
So a couple of paths i have shown here which are to be abandonned and the solution to be sought in the video provided here.
Peeping at it now I perceive that I forgot that construction can be extended beyond the given shape, and this can simplify the solution.
Thank you again, Math Booster , well done to students who " got it" on their own efforts, and encouragement and better "luck" next time to others!
Draw segments AQ & PB, and call their lengths c & d, respectively. ∠APB & ∠AQB are right angles (By Thales Thm).
So, applying Pythagorean Thm to ∆AQB & ∆APB: c² + 49 = x² and 9 + d² = x², so that c=√(x²-49) and d=√(x²-9).
All the vertices of quadrilateral APQB are on the semicircle (&, hence, the circle), so APQB is a CYCLIC quadrilateral.
So, the product of the diagonals is equal to the sum of the products of opposite sides (Ptolemy's Thm). Applying, we get:
(AQ)(PB) = (AP)(QB) + (PQ)(AB)
c d = (3)(7) + (3)x = 3(x+7)
Subbing in c and d from earlier:
√(x²-49) √(x²-9) = 3(x+7)
Squaring both sides and solving for x:
(x²-49)(x²-9) = 9(x+7)²
(x-7)(x+7)(x²-9) = 9(x+7)²
[x is a diameter which is the longest chord in the circle, so x > 7. Hence x-7 > 0. And because x-7 is not zero, we can divide both sides by it.]
(x-7)(x²-9) = 9(x+7)
x³ - 7x² - 9x + 63 = 9x + 63
x³ - 7x² - 18x = 0
x(x² - 7x -18) = 0
[x > 7 ⇒ x ≠ 0]
x² - 7x -18 = 0
(x-9)(x+2) = 0
x = 9 or x = -2 (elim., since x>7)
so x=9.
Done!!
*_Solução:_*
No triângulo retângulo ∆APB:
*sen θ = 3/x*
No triângulo retângulo AQB:
*cos 2θ = 7/x*
Como cos 2θ = 1 - 2sen² θ, então:
7/x = 1 - 2 × (3/x)²
Seja 1/x = y > 0. Daí,
7y = 1 - 18y² → 18y² + 7y - 1 = 0.
∆ = (-7)² - 4× (18)×(-1) = 121
y = (-7 + 11)/36 = 4/36 = 1/9
x = 1/y → *x = 9 unidades*
You can simplify this problem by rearranging the line segments with the 7 long one at the center. Then symmetry gives 2 triangles: one hypotenuse 3, one leg r-3.5, & the other hypotenuse r, one leg 3.5. Equating Pythagoras for the 2 other legs gives equation
3² - (r - 3.5)² = r² - (3.5)². 2r² - 7r - 9 = 0. r = 4.5. X = 2r = 9.
That is a new and useful approach . Well done.
*_Solução:_*
No triângulo retângulo ∆APQ:
*sen θ = 3/x*
No triângulo retângulo ∆AQB:
*cos 2θ = 7/x*
Como cos 2θ = 1 - 2sen² θ, então
7/x = 1 - 2 × (3/x)²
Seja 1/x = y > 0. Daí,
7y = 1 - 2 × 9y² → 18y² + 7y - 1 = 0.
∆ = (-7)² - 4×(-1)×18 = 121
y = (-7 + √121)/36 = (-7 + 11)/36
y = 4/36 = 1/9. Portanto,
1/x = y → x = 1/y = 1/(1/9)
*x = 9 unidades*
@@imetroangola17 Bueno.
I don't understand how I can simplify that problem.
@@marcusdecarvalho1354 Re-arranging the order of the line segments does not affect the radius, so they can be re-arranged to exploit symmetry to simplify the problem.
円の中心（ABの中点）を点Oとすると
△OBP∽△PQA（∵△OBP、△PQAは共に二等辺三角形、∠PQA=∠PBA（弧PAに対する円周角））
よってPA:AQ=OB:BP →PA×BP＝AQ×OB
PA＝3、AQ=√(x²-7²)、OB＝x/2、BP＝√(x²-3²)
3×√(x²-3²)=√(x²-7²)× x/2
9x²-81=(x⁴-49x²)/4
x⁴-85x²+324=0 →x²=4、81 →x=2、9
x＞7より、x=9
Taking perpendiculars from the centres of the chords and forming right triangles
2 arcsin(3/x) + arcsin(7/x) = pi/2
2 arcsin(3/x) = pi/2 - arcsin(7/x)
Take sine of both sides
2 (3/x) sqrt(1-9/x^2) = sqrt(1 - 49/x^2)
36/x^2 (1-9/x^2) = 1 - 49/x^2
Multiply by x^4, rearrange and simplify
x^4 - 85x^2 + 36.9 = 0
x^2 = (85 +/- sqrt(85^2 - 36^2)) / 2
x^2 = (85 +/- 77) / 2
x^2 = 81 or 4
x>=7 so
x=9
This looks good , efficient and neat . ( I lose the plot somewhere in line four or five. )
@@kateknowles8055 For example, in the third line of working, by the double angle formula for sine
sin(2 arcsin(3/x)) = 2 sin(arcsin(3/x)) cos(arcsin(3/x))
The sin and arcsin cancel out so the first term on the right is just 3/x. The second term is cos of something that 3/x is the sin of, so using cos^2 = 1 - sin^2 you get
sqrt(1 - 9/x^2).
With sin(pi/2 - arcsin(7/x)), you have the complementary angle so it is
cos(arcsin(7/x)) and proceed similarly.
We use an orthonormal center O, the center of the circle, and first axis (OA).
We have A(R; 0) B(-R; 0) P(R.cos(t); R.sin(t)) Q(R.cos(2.t); R.sin(2.t)) with R the radius of the circle and t = angleOAP, 0°
My solution, same as video.
Excellent !!!
Congratulations and best wishes
Or, instead......
1) Connect AQ.
2) Drop perpendicular from P to AQ, intersecting AQ at R.
3) Extend PR through O to intersect circle at S, making PS a diameter (length X), with PR = a, and RS = X - a.
4) For Right△APR, calculate (by Pythagoras) the length of AR as √(3² - a²) = √(9 - a²).
5) Use Intersecting Chords (AQ and PS) to figure 'a' in terms of X... i.e. AR * RQ = PR * RS:
√(9 - a²) * √(9 - a²) = a * (X - a), or
9 - a² = aX - a², so that
9/X = a
6) Next, use Right △AQB and Pythagoras to establish that:
7² + [2 * √(9 - a²)]² = X², which is,
49 + 36 - 4a² = X²; simplified to 'a' in terms of X becomes:
√[(85 - X²)/4] = a
7) Substituting for 'a' from #5, we get:
√[(85 - X²)/4] = (9/X), square both sides to get
(85 - X²)/4 = 81/X², or
85X² - X⁴ = 324, or
X⁴ - 85X² + 324 = 0
8) Substituting Z = X² we have:
Z² - 85Z + 324.
9) Use the quadratic equation to solve for Z:
Z = [85 ± √(85² - 4 * 1 * 324)]/2, or
Z = [85 ± √(7225 - 1296)]/2
Z = [85 ± √(5929)]/2 = [85 ± 77]/2, or
Z = 81 or 4
10) Reverting with Z = X² we get:
X = √81 = 9, or X = √4 = 2
Obviously, X cannot be equal to 2, as the resulting △AQB would be degenerate.
Therefore, X = 9. Q.E.D.
similar triangles: (a+7)/(6) = (3)/(a)
a=2
x=a+7 = 2+7=9
The answer is 14. Use the properties of cyclic quadrilaterals and the thales theorem to find out by connecting A to Q.
Nice solution! I was stucked ☺
❤
By Ptolemy's theorem
3x+21 = sqrt {(x^2 - 49)(x^2 - 9) }
x^3 - 67x - 126 = 0
X= 9 by RRT
x^3 -7•x^2 -18x=0, x>0 so x^2-7•x-18=0 so (x+2)•(x-9)=0 so x=-20 Accepted.
AQ= √3²+3²=√18=3√2
X=√ (3√2)²+7²
X=√9x2+49
X=√67
X=√9*7
X=3√7
Hay algo que no me cuadra. El problema planteado tiene infinitas soluciones si se resuelve con un compás y una regla, dependiendo del ángulo que arbitrariamente formemos entre los segmentos PA y PQ y, por ende, de la longitud del segmento AQ. Cuando un problema geométrico tiene una 'única' solución trigonométrica, tiene tambíen una 'única' solución gráfica. Y no es el caso.
x=9
🎉🎉🎉 A questão é muito boa. Eu resolvi aplicando o Teorema dos Cossenos e a minha solução não exige uma construção auxiliar. Parabéns pela escolha 🎉🎉🎉 BRASIL Novembro de 2024.
Tbm resolvi usando a Lei dos cossenos. 👍
That was a really good question
APQ=α...teorema del coseno 18-18cosα=x^2+49+14xcosα..x^2+31+(14x+18)cosα=0...poi,teorema del seno √(x^2-49)/sinα=3/sin(90-α/2)=3/cosα/2..(x^2-49)/(1-(cosα)^2)=9/(1+cosα)/2….semplifico risulta cosα=(67-x^2)/18..sostituisco mi risulta una cubica 7x^3-469x-882=0,con soluzione x=9
Que pedo pq no hay nadie que hable español xd
Applying Ptolemys Theorem we have 3•x+3•7=√(x^2-7^2)•√(x^2-3^2) so [3•(x+7)]^2=(x+7)•(x-7)•(x^2-9) , divide Both Sides by x+7>0, hence 9•(x+7)=(x-7)(x^-9) so 9x+63= x^3-9x-7x^2+63 so x^3-7x^2-18x=0,x>0 so x^2-7x-18=0 so (x+2)•(x-9)=0 so x=-2,x>0 Rejected ,or x=9 Accepted.
Nice! φ = 30°; AO = BO = PO = QO = r → 2r = AB = x = ?
AP = PQ = 3; BQ = 7; QA = k; sin(BQA) = 1 = sin(BPA)
BQO = OBQ = AOP = POQ = δ → AOQ = 2δ; ∆ ABQ → k^2 = 4r^2 - 49; cos(δ) = 7/2r
∆ AOP → 9 = 2r^2(1 - cos(δ)) = 2r^2(1 - 7/2r) → (r - 7/4)^2 = (11/4)^2 → r > 0 → r = 9/2 → x = 9 →
cos(δ) = 7/9; k = QA = 4√2; QPA = γ → cos(γ) = -7/9 → γ = 6φ - δ; AOQ = 2δ → cos(2δ) = 17/81
X=2(9/2)=9
9
asnwer=11cm isit
asnwer=9cm isit
Индийский английский противно Слушать, дизлайк за это!!
Answer is 11.27 took 1.5 minutes