Thank you, Math Booster. I got my answer quickly (by climbing a long and difficult path and then seeing a much better way. X/7 = cosine (theta) ## tan (theta) = t where t.t + 1 = 1/ cos^2(theta) = sec^2 (theta) = 49/(X.X) t.t = 49/(X.X) -1 ~~ Extending both BC and AN to meet at E, there are two simlar triangles EBA and ECN. Letting a. CN = X a .CE = BE and a..EN = EA ## 9 + EN = a.EN (a-1).EN =9 ~~ (a-1) .CE = BC and a.CN = X tan (2.theta)= BE/ X ## Angle AND is also 2.theta , because ND is parallel with BA. ~~ X-NC = ND and (a-1)CE = AD tan (2.theta) = AD/ND so BE/X = AD/ND X= BE.ND/AD = a.CE tan (2.theta) ## tan(2.theta) = 2t/ (1-t.t) ( It is positive and t
extend AM to point Q triangle ANQ is isosceles triangle AM=MQ then MN is a height AMN is a right triangle cos@=7/9=x/7 x=49/9(solved) thank u sir for this amazing video
Not so amazing vídeo!!! He looked for the most complicated solution What is really amazing are the subscribers who propose much simpler and more elegant solutions!!!!
Very beautiful puzzle here, I was able to finish most of the steps. the only part I missed was connecting MN, but I was able to continue from there and show the congruence and find X
Here's a hard way to solve the problem, but it does not use constructions! tan(Θ) = MB/BA = a/x. However, x² + a² = 7², so x = √(49 - a²) and tan(Θ) = a/(√(49 - a²)).
sin(phi) = a / 7; cos(phi) = x / 7; sin(2 * phi) = 2.sin(phi).cos(phi) = 2 * a / 7 * x / 7 = cos(90 - 2 * phi) = 2*a / 9 => x = 7*7/9 = 49/9
I also got it with: 7sin(phi)=a and 9sin(2phi)=2a; 7sin(phi)=9sin(phi)cos(phi); cos (phi)=7/9; x=7*7/9
Thank you, Math Booster.
I got my answer quickly (by climbing a long and difficult path and then seeing a much better way.
X/7 = cosine (theta)
## tan (theta) = t where t.t + 1 = 1/ cos^2(theta) = sec^2 (theta) = 49/(X.X) t.t = 49/(X.X) -1
~~ Extending both BC and AN to meet at E, there are two simlar triangles EBA and ECN. Letting a. CN = X a .CE = BE and a..EN = EA
## 9 + EN = a.EN (a-1).EN =9
~~ (a-1) .CE = BC and a.CN = X tan (2.theta)= BE/ X
## Angle AND is also 2.theta , because ND is parallel with BA.
~~ X-NC = ND and (a-1)CE = AD tan (2.theta) = AD/ND so BE/X = AD/ND X= BE.ND/AD = a.CE tan (2.theta)
## tan(2.theta) = 2t/ (1-t.t) ( It is positive and t
x=7cosθ...a=7sinθ...2a=9cos(90-2θ)=9sin2θ..a=9sinθcosθ=7sinθ=>cosθ=7/9=>x=7*7/9=49/9
extend AM to point Q
triangle ANQ is isosceles triangle
AM=MQ then MN is a height
AMN is a right triangle
cos@=7/9=x/7
x=49/9(solved)
thank u sir for this amazing video
Not so amazing vídeo!!!
He looked for the most complicated solution
What is really amazing are the subscribers who propose much simpler and more elegant solutions!!!!
Very beautiful puzzle here, I was able to finish most of the steps. the only part I missed was connecting MN, but I was able to continue from there and show the congruence and find X
2*7*cos(90°-θ) = 9 cos(90°-2θ)
14 sin θ = 9 sin 2θ
14/9 sinθ = 2 sinθ cosθ
cosθ = 7/9
x = 7 cosθ= 49/9= 5,444 cm ( Solved √ )
At 3:47, MNP is similar to AMP so a/x = (9-x)/a, and this gives a^2 + a^2 =9x.
But in AMP, a^2 + x^2 = 49, so 9x = 49, and there is your answer.
∠PMN =∠CMN = θ
and ∠AMB =∠AMP = 90-θ
Hence ∠AMP + ∠PMN =90°,
making ∆AMN a Right triangle Similar to ∆ABM.
X/7=7/9
X=49/9
Angle AMN=90° because MB=MC
Similarity of right triangles:
x/7=7/9 --> x= 49/9 cm (Solved √ )
Triangles ABM and AMN are right triangles, so cos(θ)=AB/AM=AM/AN, i.e. x/7=7/9, so x=49/9.
∠AND=2θ, ∠MNP=∠MNC=90°-θ. ∠MAN=θ. Тогда ∠AMN=90°.
По теореме Пифагора, тогда MN=4√2. И 7^(2) - x^(2)= (4√2)^(2) - (9-x)^(2).
Находим x. x=49/9
Here's a hard way to solve the problem, but it does not use constructions! tan(Θ) = MB/BA = a/x. However, x² + a² = 7², so x = √(49 - a²) and tan(Θ) = a/(√(49 - a²)).
Sinc=a/x cosc=a/7 sin2c=2a/9. Sin2c=2sinc cosc=2* a/7 * x/7=2a/9.
X=49/9
شكرا لكم
(7)^2 (9)^2={49+81}=130 {90°A+90°B+90°C+90°D}=360°ABCD/130=2.100ABCD 2^10^10 2.2^5^2^5 1.1^1^2^1 2^1 (ABCD ➖ 2ABCD+1).
This is not readable.
Easier and simpler to use trigonometry to find x.
49/9
Demorei mas acertei !