I appreciate your methods : I probably wouldn't have found either. Still, I think you could also say: sin(t) = (b/2)/5 = b/10, (t = theta), Eq. 1 sin(2t) = b/6 Eq.2 [sin(2t)]/sin(t) = [2sin(t)cos(t)]/sin(t) = 2cos(t) =[ b/6]/[b/10] = 5/3, so cos(t) = 5/6, then a = 5(5/6) = 25/6, and (b/2)^2 = 5^2 - (25/6)^2 =(900 - 625)/36, = 275/36 , so b/2 = [5{sqrt(11)}]/6, Area = ab = (25/6)[5{sqrt(11)}/3] = [125 sqrt(11)]/18
I think that the problem will be solved much faster if you drag the parallel of BC from E. Then using the same equations you' ll come up to ab=125(sqrt(11))/18!
Even if in my opinion the trigonometric solution (BC = 6*sin 2theta, BC/2 = 5*sin theta) is better, here it is a non trig. solution: If from point F we trace the perpendicular FG to line DE we get two congruent right triangles: DFC and DFG for ASA criterion. Then : FC = FG = X DG = DC If we draw a circle with center in F and radius = FC, this circle will be tangent to line DE in point G and for the two tangent theorem GE = BE = Y Now DG + GE = DE DG + Y = 6 => DG = DC = 6 - Y So we are able to write two pythagorean identities: On DFC: 5² = X² + (6 - Y)² => X² = 12Y - Y² - 11 On DEH (EH being the perpendicular to line DC) 6² = (2X)² + (6 - Y - Y)² => 4X² +4Y² - 24Y = 0 substituting the first equation in the second 48Y - 4Y² - 44 + 4Y² - 24Y = 0 Y = 11/6 DC = 6 - 11/6 = 25/6 X = (5/6)√ 11 BC = 2*5/6√ 11 = 5/3√ 11 Area rectangle = 25/6 * (5/3)√ 11 = (125/18)√ 11
Если продлить EF до пересечения с продолжением DC (пусть это точка P) то треугольник EDP равнобедренный - в нем DF биссектриса и медиана одновременно (очевидно, что EF = FP). То есть DF перпендикулярно EF, треугольники DEF и DFC подобны, откуда сразу находится DC, а затем FC, что полностью решает задачу. Что-то из этого есть в ролике, но там проще.
Good morning l think ,can be like this also Sin*=b/10 where. * is theta Cos*=a/5 Sin2*=b/6 2Sin*Cos*=b/6 2b/10×a/5=b/6 a/25=1/6 a=25/6 By Pythagoras theorem b^2/4+625/36=25 b^2/4=25-625/36 b^2/4=(900-625)/36 then we get and at the last area=about is it
Try simpler: sin∆=x/5 and sin(2∆)=2x/6=x/3 sin(2∆)=2sin∆cos∆ Therefore: 2(x/5)cos∆=x/3 x and x cancel out, so cos∆=5/6 and ∆=33.9° Knowing that you may find directly 2x and a.
3rd method FC = 5Sin@ (@=theta) therefore BC = 10sin@ by the way BC also = 6sin2@ so equation 10sin@ = 6sin2@ and sin2@ = 2sin@cos@ so 10sin@ = 12sin@cos@ there fore cos@ = 5/6 sin@ = root11/6 answer = 5cos@ x 10sin@ = 250 root11 / 36
Extend EF and let P the intersection point of EF and DC. Obviously orthogonal triangles BEF, FCP are equals => EF=FP In triangle DEP : DF is median and angle bisector at the same time => DEP is isosceles => DF is height , so EF²=ED²-DF²=36-25=> EF²=11 Let CF=x=BF , CP=y so DC =6-y x²+y²=FP² => x²+y²=11 (1) In orthogonal triangle DFP , FC is heigh => FC²=DC*CP => x²=(6-y)y => x²+y²=6y => 6y=11 , cause (1) . So y=11/6 I can easily estimate x from (1) => x=(5√11)/6 So area of ABCD = DC*BC=(6-y)2x= …….. =(125√11)/18
Let AB = x and DA = y. From the diagram we can see the following: sin(θ) = (y/2)/5 = y/10 cos(θ) = x/5 sin(2θ) = y/6 But we can also derive sin(2θ) from sin(θ) and cos(θ). sin(2θ) = 2sin(θ)cos(θ) y/6 = 2(y/10)(x/5) y/6 = 2xy/50 = xy/25 y(1/6) = y(x/25) x/25 = 1/6 x = 25/6 And we can determine y from the formula relating cos²(θ) and sin²(θ): cos²(θ) + sin²(θ) = 1 (x/5)² + (y/10)² = 1 y²/100 = 1 - (5/6)² = 1 - 25/36 y² = (11/36)100 = 275/9 y = √(275/9) = 5√11/3 A = xy = (25/6)(5√11/3) A = 125√11/18 ≈ 23.032
IlIndicando con t l'angolo theta, il metodo più semplice consiste nel calcolare t con l'equazione 6* sin(2t)=2*5*sin (t); svolgendo con le formule di duplicazione emerge per t il valore di 30° per cui b vale 5 ed a= 5*cos (t) = 25/6; poiché 5*25/6 è diverso dal valore calcolato dall'espositore ci deve essere qualche errore di calcolo.
cos(t) = 5/6 , a = 25/6 sin(t) = {sqrt(11)}/6, b = 2*5* {sqrt(11)}/6. I don't know where the 'valore di 30 [degrees]' is from, but I see that cos 30 =[sqrt(3)]/2 is close to 5/6; arccos(5/6) = 33.5573 degrees approx.(according to calculator)
I appreciate your methods : I probably wouldn't have found either. Still, I think you could also say:
sin(t) = (b/2)/5 = b/10, (t = theta), Eq. 1
sin(2t) = b/6 Eq.2
[sin(2t)]/sin(t) = [2sin(t)cos(t)]/sin(t) = 2cos(t) =[ b/6]/[b/10] = 5/3, so cos(t) = 5/6,
then a = 5(5/6) = 25/6, and
(b/2)^2 = 5^2 - (25/6)^2 =(900 - 625)/36, = 275/36 , so b/2 = [5{sqrt(11)}]/6,
Area = ab = (25/6)[5{sqrt(11)}/3] = [125 sqrt(11)]/18
I think that the problem will be solved much faster if you drag the parallel of BC from E. Then using the same equations you' ll come up to ab=125(sqrt(11))/18!
Even if in my opinion the trigonometric solution (BC = 6*sin 2theta, BC/2 = 5*sin theta) is better, here it is a non trig. solution:
If from point F we trace the perpendicular FG to line DE we get two congruent right triangles: DFC and DFG for ASA criterion. Then :
FC = FG = X
DG = DC
If we draw a circle with center in F and radius = FC, this circle will be tangent to line DE in point G and for the two tangent theorem GE = BE = Y
Now DG + GE = DE
DG + Y = 6 => DG = DC = 6 - Y
So we are able to write two pythagorean identities:
On DFC:
5² = X² + (6 - Y)² => X² = 12Y - Y² - 11
On DEH (EH being the perpendicular to line DC)
6² = (2X)² + (6 - Y - Y)² => 4X² +4Y² - 24Y = 0
substituting the first equation in the second
48Y - 4Y² - 44 + 4Y² - 24Y = 0
Y = 11/6
DC = 6 - 11/6 = 25/6
X = (5/6)√ 11
BC = 2*5/6√ 11 = 5/3√ 11
Area rectangle = 25/6 * (5/3)√ 11 = (125/18)√ 11
Если продлить EF до пересечения с продолжением DC (пусть это точка P) то треугольник EDP равнобедренный - в нем DF биссектриса и медиана одновременно (очевидно, что EF = FP). То есть DF перпендикулярно EF, треугольники DEF и DFC подобны, откуда сразу находится DC, а затем FC, что полностью решает задачу. Что-то из этого есть в ролике, но там проще.
Good morning l think ,can be like this also
Sin*=b/10 where. * is theta
Cos*=a/5
Sin2*=b/6
2Sin*Cos*=b/6
2b/10×a/5=b/6
a/25=1/6
a=25/6
By Pythagoras theorem
b^2/4+625/36=25
b^2/4=25-625/36
b^2/4=(900-625)/36
then we get and at the last area=about is it
Try simpler: sin∆=x/5 and sin(2∆)=2x/6=x/3
sin(2∆)=2sin∆cos∆
Therefore: 2(x/5)cos∆=x/3
x and x cancel out, so cos∆=5/6 and ∆=33.9°
Knowing that you may find directly 2x and a.
3rd method
FC = 5Sin@ (@=theta)
therefore
BC = 10sin@
by the way
BC also = 6sin2@
so equation 10sin@ = 6sin2@
and sin2@ = 2sin@cos@
so 10sin@ = 12sin@cos@
there fore cos@ = 5/6 sin@ = root11/6
answer = 5cos@ x 10sin@ = 250 root11 / 36
Extend EF and let P the intersection point of EF and DC.
Obviously orthogonal triangles BEF, FCP are equals => EF=FP
In triangle DEP : DF is median and angle bisector at the same time =>
DEP is isosceles => DF is height , so EF²=ED²-DF²=36-25=> EF²=11
Let CF=x=BF , CP=y so DC =6-y
x²+y²=FP² => x²+y²=11 (1)
In orthogonal triangle DFP , FC is heigh => FC²=DC*CP => x²=(6-y)y =>
x²+y²=6y => 6y=11 , cause (1) . So y=11/6
I can easily estimate x from (1) => x=(5√11)/6
So area of ABCD = DC*BC=(6-y)2x= …….. =(125√11)/18
Let AB = x and DA = y. From the diagram we can see the following:
sin(θ) = (y/2)/5 = y/10
cos(θ) = x/5
sin(2θ) = y/6
But we can also derive sin(2θ) from sin(θ) and cos(θ).
sin(2θ) = 2sin(θ)cos(θ)
y/6 = 2(y/10)(x/5)
y/6 = 2xy/50 = xy/25
y(1/6) = y(x/25)
x/25 = 1/6
x = 25/6
And we can determine y from the formula relating cos²(θ) and sin²(θ):
cos²(θ) + sin²(θ) = 1
(x/5)² + (y/10)² = 1
y²/100 = 1 - (5/6)² = 1 - 25/36
y² = (11/36)100 = 275/9
y = √(275/9) = 5√11/3
A = xy = (25/6)(5√11/3)
A = 125√11/18 ≈ 23.032
Square....right angel triangle form by drawing lines...there after triplets...then sin area formula
La clef de tous ce chemin ... est la résolution du système de deux équations avec deux inconnus...b et téta
Sin(2 téta) = b/6
Sin(téta)=b/10
Good explanation sir
IlIndicando con t l'angolo theta, il metodo più semplice consiste nel calcolare t con l'equazione 6* sin(2t)=2*5*sin (t); svolgendo con le formule di duplicazione emerge per t il valore di 30° per cui b vale 5 ed a= 5*cos (t) = 25/6; poiché 5*25/6 è diverso dal valore calcolato dall'espositore ci deve essere qualche errore di calcolo.
cos(t) = 5/6 , a = 25/6
sin(t) = {sqrt(11)}/6, b = 2*5* {sqrt(11)}/6.
I don't know where the 'valore di 30 [degrees]' is from, but I see that cos 30 =[sqrt(3)]/2 is close to 5/6;
arccos(5/6) = 33.5573 degrees approx.(according to calculator)
(250/36)vʼ11=(125/18)/vʼ11 !!!
S = (5 * cos(th)) * (2 * 5 * sin(th))
Hi sir.I would like to ask if there is any way to send math problems to you?I hope our self-created questions can be published or discussed.
You can ask the question in the comment section. I will definitely solve the problem, if the problem have good thumbnail.
What if I have some geometry questions to ask?I think this is difficult to express through just typing.
24,095 degrees!
125*sqrt(11)/18