In ∆ PSR : PR /PS = sin(angle PSR)/sin(angle SRP) = sin (angle RPQ)/sin (angle SRP) (PS being transverse to PQ || SR) = sin (3x) / sin ( π - 4 x) Again in ∆ PQR : PQ /PS = sin (angle PSQ)/sin (angle PQS) = sin ( 5 x) / sin ( π - 8 x) Herein PS = PQ Hereby sin (3x) / sin ( 4 x) = sin (5 x) / sin ( 8 x) Bit of work gives 2 sin (3 x) cos (4x) = sin (5x) sin (7x) - sin (x) = sin (5x) 2 sin(x) cos (6x) = sin(x) cos(6x) = 1/2 = cos (π/3) x = π /(18) SQ transverses PQ || RS angle QSP = angle SQP = angle QSP QS bisects angle PSQ and N be a point onto SP st SR = SN. QS is common side to both ∆ SNQ and ∆ SRQ. Hereby they are congruent. Hereby NQ = RQ angle NQS = angle RQS. angle QPN = angle NPQ = (π - angle NPQ) /2 = π/2 - 2x Again angle RQS = π - ( 8x + π/2 - 2 x) = π/2 - 6 x Again angle RQP = π - 8x
There is another geometrical solution. From point P draw segment PT parallel and equal to QR and form a parallelogram PQRT. Now notice that triangle PST is isosceles with base angle 4x. Now on side PS put a point U so that triangle PUQ is a congruent isosceles triangle with base angle 4x. Now connect points U and R to create another isosceles triangle with base angles 2x. Now notice that triangle PUR is another isosceles triangle with base angle x. From all these you can conclude that triangle QRU is equilateral with one angle equal to 6x.
if PS=PQ=1 the sine theorem PR=sin(180°-8x)/sin5x=sin(180°-4x)/sin3x ∴sin4xsin5x=sin8xsin3x=2sin4xcos4xsin3x ∴sin5x=2cos4xsin3x=sin7x-sinx ∴sinx=sin7x-sin5x=2cos6xsinx ∴cos6x=1/2 6x=60° ∴x=10°
Your solution seems to skip a lot of steps, and you keep forgetting to include "sin" For example, in sinx/(2cos4x sin4x) = 3x/4x, right side should be sin3x/sin4x Also, theta should be pi
In ∆ PSR :
PR /PS
= sin(angle PSR)/sin(angle SRP)
= sin (angle RPQ)/sin (angle SRP)
(PS being transverse to PQ || SR)
= sin (3x) / sin ( π - 4 x)
Again in ∆ PQR :
PQ /PS
= sin (angle PSQ)/sin (angle PQS)
= sin ( 5 x) / sin ( π - 8 x)
Herein PS = PQ
Hereby
sin (3x) / sin ( 4 x)
= sin (5 x) / sin ( 8 x)
Bit of work gives
2 sin (3 x) cos (4x)
= sin (5x)
sin (7x) - sin (x) = sin (5x)
2 sin(x) cos (6x) = sin(x)
cos(6x) = 1/2 = cos (π/3)
x = π /(18)
SQ transverses PQ || RS
angle QSP = angle SQP
= angle QSP
QS bisects angle PSQ and
N be a point onto SP st SR = SN.
QS is common side to both
∆ SNQ and ∆ SRQ. Hereby they are congruent.
Hereby NQ = RQ
angle NQS = angle RQS.
angle QPN = angle NPQ
= (π - angle NPQ) /2
= π/2 - 2x
Again angle RQS
= π - ( 8x + π/2 - 2 x) = π/2 - 6 x
Again angle RQP
= π - 8x
There is another geometrical solution. From point P draw segment PT parallel and equal to QR and form a parallelogram PQRT. Now notice that triangle PST is isosceles with base angle 4x. Now on side PS put a point U so that triangle PUQ is a congruent isosceles triangle with base angle 4x. Now connect points U and R to create another isosceles triangle with base angles 2x. Now notice that triangle PUR is another isosceles triangle with base angle x. From all these you can conclude that triangle QRU is equilateral with one angle equal to 6x.
if PS=PQ=1 the sine theorem PR=sin(180°-8x)/sin5x=sin(180°-4x)/sin3x
∴sin4xsin5x=sin8xsin3x=2sin4xcos4xsin3x
∴sin5x=2cos4xsin3x=sin7x-sinx
∴sinx=sin7x-sin5x=2cos6xsinx ∴cos6x=1/2 6x=60° ∴x=10°
Əla həll etdiniz.Təşəkkürlər.Bakıdan salamlar.
Your problem has many solutions from degree 0 to 14, precisely in the interval ]0, 14[, so it's possible to have x different from 10.
Can you please elaborate on that. Because from what I got using the Law of Sines (her method #1 in the video), the only possible solution is x = 10°.
good one
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لدينا ما لا نهاية له من الحلول من 0 إلى 14 درجة اي ]0,14[
If PQ=PS PQ||SR
Your solution seems to skip a lot of steps, and you keep forgetting to include "sin"
For example, in sinx/(2cos4x sin4x) = 3x/4x, right side should be sin3x/sin4x
Also, theta should be pi
@@MarieAnne. Thanks goodness
Using triangle PSR to confirm the answer x+(Pi-4x)+3x= Pi, I discover that it works for any values of x... What does that mean ?
S=]0,14[
X peut être différent de10 degrés
X peut prendre toutes les valeurs de 0 à 14 degrés S=]0,14[
شكرا لكم
الحلول هي كل عدد x من المجال ]0,14[=S
Votre probleme a beaucoup de solution de 0 degré à 14 degrés plus précisément S=]0,14[ on peut avoir x différent de 10
S=]0,14[ degres