A possible non trigonometric solution can be this: Draw a quarter circle in the low side of the figure getting a semicircle with center O and diameter AD Extend radius PC untill it intersects diameter of semicircle in point Q, so we have 3 similar right triangle, ABD, ACQ and OPQ. AC = 8 (two tangent theorem), PC = 4, OQ = x, PQ = y considering triangle ACQ and OPQ AC:OP = AQ:PQ 8:4=(8+X):Y X - 2Y + 8 = 0 (1) AC:OP= CQ: OQ 8:4= (4+Y):X 2X -Y -4 = 0 (2) Solving the system (1)+(2) OQ=x = 16/3, PQ=y = 20/3 AQ = OA+OQ = 8+16/3= 40/3 considering triangle ABD and ACQ: AC : AB = AQ : AD 8 : AB = 40/3 : (8+8) AB = 48/5
At 4:00, from ΔAPO, we note that tan(Θ) = 4/8 = 1/2. Applying the tangent double angle formula, tan(2Θ) = 4/3. Extend the quarter circle into a full circle. Extend radius AO into a diameter, labelling the other end of the diameter as point D. Construct BD.
let the small circle be (x-4)^2 + y^2 = 4^2 the tangent line passes through (0,8) to the non y-axis surface the circle. let tangent line be y=mx+b, find b, subtitute (0,8) in y=mx+b 8= (m*0) + b b=8 perpendicular distance formula r = |A*x1 + B*y1 + C| / (sqrt(A^2+B^2)) r= radius of the small circle rearrage y=mx+8 to mx-y+8=0 A=m, B=-1 (coefficent of y), C=8 sub A, B, C and (4,0) the center of the small circle in |m*x1 + B*y1 + b| / (sqrt(A^2+B^2)) |m*4 + 0*-1 + b| / (sqrt(m^2+(-1)^2)) = |4m+b|/sqrt(m^2+1) = R r=4 radius of the small circle |4m+b|/sqrt(m^2+1) = 4 find m, m = -3/4 tangent line = y=-3/4x + 8 let the big circle be x^2+y^2=8^2 get intersections of x^2+y^2=8^2 and y=-3/4x + 8 intersections are: (0,8) and (192/25 , 56/25) get distance between (0,8) and (192/25 , 56/25) d=sqrt((192/25 - 0)^2+(56/25-8)^2) = 48/5 = 9.6
I just happened to stumble across this video… for context I HATE MATH(maybe because I never understood it).. geometry was the only thing I could grasp in high school. Listening sounds as if he’s speaking in a different language. Watching just looks like scrambles on the screen. After about two minutes I was completely lost. Going through the comments and seeing everyone fully engaged is intriguing. I envy you guys😩.. I wished my mind was amazing like that
Let's use an orthonormal center O and first axis (OP). The equation of the little half circle is (x -4)^2 + y^2 = 16 or x^2 + y^2 -8.x = 0 The equation of (OB) is y -8 = m.x (m unknown real). At the intersection C we have: x^2 + (m.x +8)^2 -8.x = 0 or (1+m^2).(x^2) + (16.m -8).x +64 = 0 C (point of tangency) is a "double" point of intersection of (OB) and the little circle, so this equation has a double solution, its discriminant is 0 Deltaprime = (8.m -4)^2 -64.(1 +m^2) = 64.(m^2) -64.m +16 -64 -64.(m^2) = -64.m -48 = 0, so m = -48/64 = -3/4. The equation of (OB) is then y = (-3/4).x +8 The equation of the big circle is x^2 + y^2 = 64, so at the intersection B with (OB) we have: (x^2) + ((-3/4).x +8)^2 = 64 or (25/16).(x^2) -12.x = 0 So the abscissa of B is 12/(25/16) = 192/25 (the other solution 0 is the abscissa of A). The ordinate of B is now (-3/4).(192/25) + 8 = 56/25 So we have B(192/25; 56/25), and VectorAB(192/25; -144/25). Finally AB^2 = (36864/625) + (20736/625) = 57600/625 and AB = 240/25 = 48/5.
Pure geometry is fun😅 1/ AC= 8 (Tangent theorem) 2/Label M as the end of horizontal radius of the quadrant. Buid the tagent of quadrant from M intersecting AB(extended) at point P. We have PC=PM= 2->PA=10 -> sq PM=PBxPA -> 4= 10 PB -> PB=0.4 AB= 10-0.4= 9.6 units😅😅😅 3/ Alternative solution: Tan( theta)= 1/2 -> tan(2theta)= OD/AD= 4/3 -> the triangle AOD is a 3-4-5 triples So, AD/OA= 3/5 -> AD=24/5 -> AB = 2AD= 48/5= 9.6 units😅😅😅
Looking also at the various solutions proposed by other users, we can conclude that we are faced with a very solvable problem: we just have to choose the method that we like the most
AC=8---> AP=√(8²+4²)=4√5---> OC=2*(8*4/4√5) =16√5/5 ---> Potencia de C respecto a la circunferencia grande. =8*CB =(8 -16√5/5)*(2*8 -8 +16√5/5)---> CB =1,60---> AB =8+1,60=9,60. Gracias y saludos
*Solução:* Com respeito a circunferência maior ou um quarto do círculo maior, o ponto C é um ponto interior ao círculo maior, usando o teorema de potência num ponto interior ao círculo em relação ao ponto C, teremos: AC × BC = R² - OC², onde R é o raio do círculo maior. Como AC e AO são retas tangentes ao círculo menor (ou o semicírculo menor), vamos ter que AC=AO=8. Além disso, R=AO =8. Vamos calcular OC. Observe que o ∆AOC é isósceles, se traçamos a altura no ponto T pertencente ao segmento OC, teremos OT é mediana e bissetriz, já que o ∆AOC é isósceles. Por outro lado, o ∆OCP é isósceles, pois PC=OP que é o raio do círculo menor, logo o segmento PT também é altura, medianas e bissetriz do ∆OCP. Consequentemente, os triângulos ∆APC (retângulo em C) e o ∆ATC são semelhantes, assim: *TC/PC = AC/AP* PC é o raio do círculo menor, donde PC = 8/2 = 4. Ora, como o triângulo ∆AOP é retângulo, então: AP² = AO² + OP² = 8² + 4² = 80 AP= √80 = 4√5. Daí, TC/4 = 8/4√5 →TC = 8/√5 e, OC = 2×TC → OC = 16/√5. Lembrando que: *AC × BC = R² - OC²* Assim, 8 × BC = 8² - (16/5)² 8 × BC = 8² - 16²/5 ÷ (8) BC = 8 - 32/5 = (40 - 32)/5 = 8/5 BC = 1,6. Concluímos que: AB = AC + BC = 8 + 1,6 *AB = 9,6 unidades.* Professor gostei bastante de sua solução! Vou deixar a minha solução para abrir mais nossos horizontes! Abraços!
This might be good practice to remember all of the relevant circle theorems so that you can train yourself in making the correct visualization and pair that to the correct algebraic manipulation. The anser is 9.6 units.
Nice Congruency of AOP and ACP may be proved with this logic They r rt triangles. Hypotenuse is common. Another sides OP =PC = radius of the small semicircle.
STEP-BY-STEP RESOLUTION PROPOSAL : 01) Equation of the Straigth Line AB : Y = 3X/4 + 8 02) Equation ot the Big Circle: X^2 + Y^2 = 64 03) Coordinates of Intersetion Point (B) : X = 192/25 or X = 7,68 ; Y = 56/25 or Y = 2,24 04) Equation of the Straigth Line AB : Y = 3X/4 + 8 05) Equation ot the Small Circle : (X - 4)^2 + Y^2 = 16 06) Coordinates of Intersetion Point (C) : X = 32/5 or X = 6,4 ; Y = 16/5 or Y = 3,2 07) Distance between Two Points : 08) d (B,C) = sqrt(1,28^2 + 0,96^2) ; d (B,C) = sqrt(1,6384 + 0,9216) ; d(B,C) = sqrt(2,56) ; d(BC) = 1,6 lin un ; BC = 1,6 lin un 09) As : AO = AC = 8 lin un 10) AB = AC + BC ; AB = (8 + 1,6) lin un ; AB = 9,6 lin un Therefore, OUR BEST ANSWER : Length of Line AB equal to 9,6 Linear Units VERY IMPORTANT NOTE: I would like not to be harassed here for Ignorant People who doesn't know what a "Linear Unit" is or a "Square Unit" is!! Thank You Very Much. I don't come here to have a "Good Time"; this is not a Party. I come here to Learn!!
I found a purely geometric approach for anyone interested here and got the exact same answer (note, I haven't read all the comments here yet so dunno if someone already wrote one before me)
@harrydon-z3j Sure, first though, I think I was able to do it in 2 kinda different ways without trig. The first was the more interesting one, in my opinion, and it went like this. Label the centres of the quarter circle and the semicircle M and N, respectively. Label the intersection of AB and circle N as C. Label the right side intersection of 🔵 M and 🔵 N as P. Extend lines AB and MN to meet on the right of the circles at point D. Let CD = a, and DP = b, using the tangent-secant circle theorem on circle N, a² = b(b + 8) and call that eq 1. In 🔺️ ADM, AC = AM = 8 [circle tangent theorem or sth on M], (8 + a)² = 8² + (8 + b)² call that eq 2. Solving eq 1 & 2 together, we find that a = 16/3, b = 8/3. Now, construct line MQ, where Q is the midpoint of chord AB, MQ is also perpendicular on AB [chord bisector theorem or sth]. IF AB = 2x, then AQ = x. 🔺️ MQA shares angle MAB with 🔺️ ADM, and they both have 90deg angles, so 🔺️ MAQ ~ 🔺️ DAM. So AQ / AM = AM / AD, x / 8 = 8 / (8 + 16/3), and x = 24/5 = 4.8, and AB = 2x = 48/5 = 9.6 As for the other method, though, I saw it in another video. You do mainly the same but apply the tangent-secant theorem to both circle N and circle M by completing circle M into a semicircle
Rifaccio i calcoli AB=AC+CB...AC=8..per il calcolo di CB,calcolo PB..per la legge del coseno 8^2=4^2+PB^2-2*4*PBcos(2arctg2+arccos4/PB)...dopo i calcoli risultano PB^2=18,56..PB^2=80...nel primo caso risulta CB^2=PB^2-16=18,56-16=2,56..CB=1,6...quindi AB=8+1,6=9,6
By the two tangent theorem, as CA and OA are both tangent to semicircle P and intersect at A, CA = OA = 8. Draw radius PC. As AB is tangent to semicircle P at C, then ∠PCA = 90°. Triangle ∆PCA: PC² + CA² = PA² 4² + 8² = PA² PA² = 16 + 64 = 80 PA = √80 = 4√5 Let ∠PAO = ∠CAP = θ. cosθ = CA/PA = 8/4√5 = 2/√5 Let M be the midpoint of AB. As AB is a chord, ∠OMA = 90°, as OM is a perpendicular bisector of AB. By observation, ∠MAO = 2θ. cos2θ = 2cos²θ - 1 MA/OA = 2(2/√5)² - 1 MA/8 = 2(4/5) - 1 MA/8 = 8/5 - 1 = 3/5 MA = (8)3/5 = 24/5 AB = 2MA = 2(24/5) = 48/5 = 9.6 units
Let's face this challenge: . .. ... .... ..... According to the two tangent theorem we know that AC=OA=8. The relation between the radii R and r of the quarter circle and the semicircle, respectively, is obviously given by r=R/2=8/2=4. The triangles OAP and PAC are right triangles (the latter one since AB is a tangent to the semicircle). Therefore we can conclude: tan(∠OAC) = tan(∠OAP + ∠PAC) = [tan(∠OAP) + tan(∠PAC)]/[1 − tan(∠OAP)tan(∠PAC)] = (OP/OA + PC/AC)/[1 − (OP/OA)(PC/AC)] = (r/R + r/R)/[1 − (r/R)(r/R)] = (1/2 + 1/2)/[1 − (1/2)*(1/2)] = 1/(1 − 1/4) = 1/(3/4) = 4/3 Let's add point D on OA such that the triangle ACD is a right triangle. Then we obtain: ∠CAD = ∠OAC ⇒ tan(∠CAD) = tan(∠OAC) ⇒ CD/AD = 4/3 ⇒ tan(∠ACD) = AD/CD = 3/4 Now let's assume that O is the center of the coordinate system and OP is located on the x-axis. Then the line ACB can be represented by the following function: y = −tan(∠ACD)*x + yA = −(3/4)*x + 8 The function representing the quarter circle is: (x − xO)² + (y − yO)² = R² (x − 0)² + (y − 0)² = 8² x² + y² = 64 Therefore the coordinates of the point of intersection (B) between the linear function and the quarter circle can be calculated in the following way: xB² + [−(3/4)*xB + 8]² = 64 xB² + (9/16)xB² − 12xB + 64 = 64 (25/16)xB² − 12xB = 0 25xB² = 192xB ⇒ xB = 0 ∨ xB = 192/25 The first solution represents point A. So the only useful solution is: xB = 192/25 yB = −(3/4)*xB + 8 = −(3/4)*(192/25) + 8 = −144/25 + 8 = −144/25 + 200/25 = 56/25 Now we are able to calculate the length of AB: AB² = (xB − xA)² + (yB − yA)² = (192/25 − 0)² + (56/25 − 8)² = (192/25)² + (56/25 − 200/25)² = (192/25)² + (−144/25)² = (192² + 144²)/25² = (48²*4² + 48²*3²)/25² = 48²*5²/25² = 48²/5² ⇒ AB = 48/5 = 9.6 Best regards from Germany
Thank you! I always enjoy the use of Algebra and Trigonometry when solving!
I'm glad you appreciate the combination of Algebra and Trigonometry! ❤️
You are very welcome!
Thanks for the feedback ❤️
A possible non trigonometric solution can be this:
Draw a quarter circle in the low side of the figure getting a semicircle with center O and diameter AD
Extend radius PC untill it intersects diameter of semicircle in point Q, so we have 3 similar right triangle, ABD, ACQ and OPQ.
AC = 8 (two tangent theorem), PC = 4, OQ = x, PQ = y
considering triangle ACQ and OPQ
AC:OP = AQ:PQ
8:4=(8+X):Y
X - 2Y + 8 = 0 (1)
AC:OP= CQ: OQ
8:4= (4+Y):X
2X -Y -4 = 0 (2)
Solving the system (1)+(2)
OQ=x = 16/3,
PQ=y = 20/3
AQ = OA+OQ = 8+16/3= 40/3
considering triangle ABD and ACQ:
AC : AB = AQ : AD
8 : AB = 40/3 : (8+8)
AB = 48/5
Very clever use of similar triangles. Bravo.
@ thank you 🙏 similarity is the essence of the euclidean geometry
At 4:00, from ΔAPO, we note that tan(Θ) = 4/8 = 1/2. Applying the tangent double angle formula, tan(2Θ) = 4/3. Extend the quarter circle into a full circle. Extend radius AO into a diameter, labelling the other end of the diameter as point D. Construct BD.
In my opinion, finding that tan ( 2 theta)= 4/3 is enough to demonstrate that the triangle AOD is a 3-4-5 one-> AD/OA =3/5😅
nice!
I wish I thought of the using the notion of the radius bisecting the chord. Thank you, PreMath, for your thoroughly worked-out simple solution.
Glad it was helpful!
You are very welcome!
Thanks for the feedback ❤️
let the small circle be (x-4)^2 + y^2 = 4^2
the tangent line passes through (0,8) to the non y-axis surface the circle.
let tangent line be y=mx+b, find b,
subtitute (0,8) in y=mx+b
8= (m*0) + b
b=8
perpendicular distance formula
r = |A*x1 + B*y1 + C| / (sqrt(A^2+B^2))
r= radius of the small circle
rearrage y=mx+8 to mx-y+8=0
A=m, B=-1 (coefficent of y), C=8
sub A, B, C and (4,0) the center of the small circle in |m*x1 + B*y1 + b| / (sqrt(A^2+B^2))
|m*4 + 0*-1 + b| / (sqrt(m^2+(-1)^2))
= |4m+b|/sqrt(m^2+1) = R
r=4 radius of the small circle
|4m+b|/sqrt(m^2+1) = 4
find m, m = -3/4
tangent line = y=-3/4x + 8
let the big circle be x^2+y^2=8^2
get intersections of x^2+y^2=8^2 and y=-3/4x + 8
intersections are: (0,8) and (192/25 , 56/25)
get distance between (0,8) and (192/25 , 56/25)
d=sqrt((192/25 - 0)^2+(56/25-8)^2)
= 48/5 = 9.6
The solution to a problem with Cartesian coordinates is always very computationally hard, but not in this case
I just happened to stumble across this video… for context I HATE MATH(maybe because I never understood it).. geometry was the only thing I could grasp in high school. Listening sounds as if he’s speaking in a different language. Watching just looks like scrambles on the screen. After about two minutes I was completely lost. Going through the comments and seeing everyone fully engaged is intriguing. I envy you guys😩.. I wished my mind was amazing like that
Beautiful problem. Thank you.
Let's use an orthonormal center O and first axis (OP). The equation of the little half circle is (x -4)^2 + y^2 = 16 or x^2 + y^2 -8.x = 0
The equation of (OB) is y -8 = m.x (m unknown real). At the intersection C we have: x^2 + (m.x +8)^2 -8.x = 0 or (1+m^2).(x^2) + (16.m -8).x +64 = 0
C (point of tangency) is a "double" point of intersection of (OB) and the little circle, so this equation has a double solution, its discriminant is 0
Deltaprime = (8.m -4)^2 -64.(1 +m^2) = 64.(m^2) -64.m +16 -64 -64.(m^2) = -64.m -48 = 0, so m = -48/64 = -3/4. The equation of (OB) is then y = (-3/4).x +8
The equation of the big circle is x^2 + y^2 = 64, so at the intersection B with (OB) we have: (x^2) + ((-3/4).x +8)^2 = 64 or (25/16).(x^2) -12.x = 0
So the abscissa of B is 12/(25/16) = 192/25 (the other solution 0 is the abscissa of A). The ordinate of B is now (-3/4).(192/25) + 8 = 56/25
So we have B(192/25; 56/25), and VectorAB(192/25; -144/25). Finally AB^2 = (36864/625) + (20736/625) = 57600/625 and AB = 240/25 = 48/5.
Pure geometry is fun😅
1/ AC= 8 (Tangent theorem)
2/Label M as the end of horizontal radius of the quadrant. Buid the tagent of quadrant from M intersecting AB(extended) at point P.
We have PC=PM= 2->PA=10
-> sq PM=PBxPA
-> 4= 10 PB
-> PB=0.4
AB= 10-0.4= 9.6 units😅😅😅
3/ Alternative solution:
Tan( theta)= 1/2 -> tan(2theta)= OD/AD= 4/3
-> the triangle AOD is a 3-4-5 triples
So, AD/OA= 3/5 -> AD=24/5
-> AB = 2AD= 48/5= 9.6 units😅😅😅
The first method is very nice! Congratulation!
How did u get value of PC, PM as 2 , plz elaborate...thanks in advance
@@harrydon-z3ja way could be with Pythagorean theorem:
(8+x)^2=8^2+(8-x)^2
With x=PC=PM
Thanks very much for prompt reply ❤
Looking also at the various solutions proposed by other users, we can conclude that we are faced with a very solvable problem: we just have to choose the method that we like the most
AC=8---> AP=√(8²+4²)=4√5---> OC=2*(8*4/4√5) =16√5/5 ---> Potencia de C respecto a la circunferencia grande. =8*CB =(8 -16√5/5)*(2*8 -8 +16√5/5)---> CB =1,60---> AB =8+1,60=9,60.
Gracias y saludos
Nice! And AOPC is a kite whose diagonals are perpendicular
@@soli9mana-soli4953 Cierto; ya decían en la Academia: "Nadie entre aquí que no sepa volar una cometa". Gracias y un saludo cordial.
@@santiagoarosam430 molto bello come monito per chiunque si avvicini alle discipline scientifiche!
*Solução:*
Com respeito a circunferência maior ou um quarto do círculo maior, o ponto C é um ponto interior ao círculo maior, usando o teorema de potência num ponto interior ao círculo em relação ao ponto C, teremos:
AC × BC = R² - OC², onde R é o raio do círculo maior.
Como AC e AO são retas tangentes ao círculo menor (ou o semicírculo menor), vamos ter que AC=AO=8. Além disso, R=AO =8. Vamos calcular OC. Observe que o ∆AOC é isósceles, se traçamos a altura no ponto T pertencente ao segmento OC, teremos OT é mediana e bissetriz, já que o ∆AOC é isósceles. Por outro lado, o ∆OCP é isósceles, pois PC=OP que é o raio do círculo menor, logo o segmento PT também é altura, medianas e bissetriz do ∆OCP. Consequentemente, os triângulos ∆APC (retângulo em C) e o ∆ATC são semelhantes, assim:
*TC/PC = AC/AP*
PC é o raio do círculo menor, donde PC = 8/2 = 4. Ora, como o triângulo ∆AOP é retângulo, então:
AP² = AO² + OP² = 8² + 4² = 80
AP= √80 = 4√5. Daí,
TC/4 = 8/4√5 →TC = 8/√5 e, OC = 2×TC → OC = 16/√5.
Lembrando que:
*AC × BC = R² - OC²*
Assim,
8 × BC = 8² - (16/5)²
8 × BC = 8² - 16²/5 ÷ (8)
BC = 8 - 32/5 = (40 - 32)/5 = 8/5
BC = 1,6. Concluímos que:
AB = AC + BC = 8 + 1,6
*AB = 9,6 unidades.*
Professor gostei bastante de sua solução! Vou deixar a minha solução para abrir mais nossos horizontes! Abraços!
圆弧AB方程：x^2+y^2=64,A B的方程：y =-3/4x+8，联立可以解出B点坐标。
This might be good practice to remember all of the relevant circle theorems so that you can train yourself in making the correct visualization and pair that to the correct algebraic manipulation. The anser is 9.6 units.
We can call circle theorem as radius tangent theorem.
Very good
Nice
Congruency of AOP and ACP may be proved with this logic
They r rt triangles. Hypotenuse is common. Another sides OP =PC = radius of the small semicircle.
Bom dia Mestre
Esse foi hot
Grato pela aula
Tan x=1/2=t, y=(180-4x)/2=90-2x, AB=2×8 sin y=16 cos 2x=16 × (1-t^2)/(1+t^2)=16× 3/5=48/5.😅
STEP-BY-STEP RESOLUTION PROPOSAL :
01) Equation of the Straigth Line AB : Y = 3X/4 + 8
02) Equation ot the Big Circle: X^2 + Y^2 = 64
03) Coordinates of Intersetion Point (B) : X = 192/25 or X = 7,68 ; Y = 56/25 or Y = 2,24
04) Equation of the Straigth Line AB : Y = 3X/4 + 8
05) Equation ot the Small Circle : (X - 4)^2 + Y^2 = 16
06) Coordinates of Intersetion Point (C) : X = 32/5 or X = 6,4 ; Y = 16/5 or Y = 3,2
07) Distance between Two Points :
08) d (B,C) = sqrt(1,28^2 + 0,96^2) ; d (B,C) = sqrt(1,6384 + 0,9216) ; d(B,C) = sqrt(2,56) ; d(BC) = 1,6 lin un ; BC = 1,6 lin un
09) As : AO = AC = 8 lin un
10) AB = AC + BC ; AB = (8 + 1,6) lin un ; AB = 9,6 lin un
Therefore,
OUR BEST ANSWER :
Length of Line AB equal to 9,6 Linear Units
VERY IMPORTANT NOTE: I would like not to be harassed here for Ignorant People who doesn't know what a "Linear Unit" is or a "Square Unit" is!!
Thank You Very Much. I don't come here to have a "Good Time"; this is not a Party.
I come here to Learn!!
1.2 R . Ok simple
I found a purely geometric approach for anyone interested here and got the exact same answer (note, I haven't read all the comments here yet so dunno if someone already wrote one before me)
Plz share pure geometric solution
@harrydon-z3j Sure, first though, I think I was able to do it in 2 kinda different ways without trig. The first was the more interesting one, in my opinion, and it went like this.
Label the centres of the quarter circle and the semicircle M and N, respectively.
Label the intersection of AB and circle N as C.
Label the right side intersection of 🔵 M and 🔵 N as P.
Extend lines AB and MN to meet on the right of the circles at point D.
Let CD = a, and DP = b, using the tangent-secant circle theorem on circle N,
a² = b(b + 8) and call that eq 1.
In 🔺️ ADM, AC = AM = 8 [circle tangent theorem or sth on M],
(8 + a)² = 8² + (8 + b)² call that eq 2.
Solving eq 1 & 2 together, we find that a = 16/3, b = 8/3.
Now, construct line MQ, where Q is the midpoint of chord AB,
MQ is also perpendicular on AB [chord bisector theorem or sth].
IF AB = 2x, then AQ = x.
🔺️ MQA shares angle MAB with 🔺️ ADM, and they both have 90deg angles, so 🔺️ MAQ ~ 🔺️ DAM.
So AQ / AM = AM / AD, x / 8 = 8 / (8 + 16/3), and x = 24/5 = 4.8, and AB = 2x = 48/5 = 9.6
As for the other method, though, I saw it in another video. You do mainly the same but apply the tangent-secant theorem to both circle N and circle M by completing circle M into a semicircle
Thanks ❤
@harrydon-z3j No problem
Always 1.2 times radius,, isn't it nice and that angle 2 theta is always be a constant 3/5
Rifaccio i calcoli AB=AC+CB...AC=8..per il calcolo di CB,calcolo PB..per la legge del coseno 8^2=4^2+PB^2-2*4*PBcos(2arctg2+arccos4/PB)...dopo i calcoli risultano PB^2=18,56..PB^2=80...nel primo caso risulta CB^2=PB^2-16=18,56-16=2,56..CB=1,6...quindi AB=8+1,6=9,6
By the two tangent theorem, as CA and OA are both tangent to semicircle P and intersect at A, CA = OA = 8.
Draw radius PC. As AB is tangent to semicircle P at C, then ∠PCA = 90°.
Triangle ∆PCA:
PC² + CA² = PA²
4² + 8² = PA²
PA² = 16 + 64 = 80
PA = √80 = 4√5
Let ∠PAO = ∠CAP = θ.
cosθ = CA/PA = 8/4√5 = 2/√5
Let M be the midpoint of AB. As AB is a chord, ∠OMA = 90°, as OM is a perpendicular bisector of AB. By observation, ∠MAO = 2θ.
cos2θ = 2cos²θ - 1
MA/OA = 2(2/√5)² - 1
MA/8 = 2(4/5) - 1
MA/8 = 8/5 - 1 = 3/5
MA = (8)3/5 = 24/5
AB = 2MA = 2(24/5) = 48/5 = 9.6 units
Angle OAB = α ; R=8cm ; c=AB
tan½α = ½R/R = 1/2 --> α=53,13°
½c = R cosα ---> c=9,6cm ( Solved√)
Too complicated video solution !!!!
Let's face this challenge:
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According to the two tangent theorem we know that AC=OA=8. The relation between the radii R and r of the quarter circle and the semicircle, respectively, is obviously given by r=R/2=8/2=4. The triangles OAP and PAC are right triangles (the latter one since AB is a tangent to the semicircle). Therefore we can conclude:
tan(∠OAC)
= tan(∠OAP + ∠PAC)
= [tan(∠OAP) + tan(∠PAC)]/[1 − tan(∠OAP)tan(∠PAC)]
= (OP/OA + PC/AC)/[1 − (OP/OA)(PC/AC)]
= (r/R + r/R)/[1 − (r/R)(r/R)]
= (1/2 + 1/2)/[1 − (1/2)*(1/2)]
= 1/(1 − 1/4)
= 1/(3/4)
= 4/3
Let's add point D on OA such that the triangle ACD is a right triangle. Then we obtain:
∠CAD = ∠OAC ⇒ tan(∠CAD) = tan(∠OAC) ⇒ CD/AD = 4/3 ⇒ tan(∠ACD) = AD/CD = 3/4
Now let's assume that O is the center of the coordinate system and OP is located on the x-axis. Then the line ACB can be represented by the following function:
y = −tan(∠ACD)*x + yA = −(3/4)*x + 8
The function representing the quarter circle is:
(x − xO)² + (y − yO)² = R²
(x − 0)² + (y − 0)² = 8²
x² + y² = 64
Therefore the coordinates of the point of intersection (B) between the linear function and the quarter circle can be calculated in the following way:
xB² + [−(3/4)*xB + 8]² = 64
xB² + (9/16)xB² − 12xB + 64 = 64
(25/16)xB² − 12xB = 0
25xB² = 192xB
⇒ xB = 0 ∨ xB = 192/25
The first solution represents point A. So the only useful solution is:
xB = 192/25
yB = −(3/4)*xB + 8 = −(3/4)*(192/25) + 8 = −144/25 + 8 = −144/25 + 200/25 = 56/25
Now we are able to calculate the length of AB:
AB²
= (xB − xA)² + (yB − yA)²
= (192/25 − 0)² + (56/25 − 8)²
= (192/25)² + (56/25 − 200/25)²
= (192/25)² + (−144/25)²
= (192² + 144²)/25²
= (48²*4² + 48²*3²)/25²
= 48²*5²/25²
= 48²/5²
⇒ AB = 48/5 = 9.6
Best regards from Germany
My answer is 10????!!!! @@""
Non è corretto... anch'io avevo calcolato 10,ma è sbagliato
I got 9.6 as answer by using tangent secant property ..