A very easy solution (it just took a few seconds to solve!) - Draw a perpendicular from point Q. QM will be half of AB( because ABC and QMC are similar). And we know AB=QP. So, Triangle PQM is a 30-60-90 Triangle. Thus, we know that angle PQM is 60°. Angle MQC is 70°. So angle theta= 180-(60+70)=50°. Hope this helps!
Let CQ = QA = a and AB = PQ = b. Draw QT, where T is the point on BC where QT is perpendicular to BC. As QT is also parallel to AB, then ∠CQT and ∠CAB are corresponding angles and thus ∠CQT = ∠CAB = 90°-20° = 70°. As ∠QTC = ∠ABC = 90° and ∠C is common, then ∆QTC and ∆ABC are similar. QT/AB = CQ/CA QT/b = a/2a = 1/2 QT = b/2 Let ∠PQT = α. cosα = QT/PQ = (b/2)/b cosα = 1/2 α = cos⁻¹(1/2) = 60° θ + α + 70° = 180° θ + 60° + 70° = 180° θ + 130° = 180° [ θ = 180° - 130° = 50° ]
Assign a value of 1 to line AC. At 20 degrees, line AB (sine of 20 degrees) will equal .3420. Line QP is thus .3420. Line QC is .5 . Going Pythagorean, consider triangle PQC. Angle at C is 20 degrees. Angle at P is 30 degrees. Angle at apex Q is 130 degrees. Line QM is .1710 . Indicated exterior angle at Q is 130-180 degrees or 50 degrees. Reasonably simple problem.
Tracing the line BQ and setting AQ=QC=x, AB=QP=y, we can write 2 identities with sines law: On ABQ Y/sin 40=x/sin 70 On PQC Y/sin 20=x/sin alpha Alpha is angle QPC Comparing we get: Sin alpha = sin 70*sin 20/sin 40 And knowing that Sin 70 = sin (90-20)=Cos 20 Sin 40=2sin 20*cos 20 Sin alpha =1/2 Alpha =30 Theta = 20+30=50
AB = PQ = a, AQ = CQ = b, ∠CPQ = α From △ABC we get: tan(20°) = a/(2b) Using law of sines in △CPQ we get sin(α) / b = sin(20°) / a sin(α) = b/a * sin(20°) = b/a * a/(2b) = 1/2 α = 30° θ is an exterior angle of △CPQ, which is equal to the sum of its two opposite angles θ = α + 20° = 30° + 20° *θ = 50°*
These problems seem ten times easier after they have been worked out! (I have deleted my false trails this time.) Angle A =70 degrees If QM is the perpendicular from Q to BC, it is (1/2) of a because triangles ABC and QMC are similar. PQM is 60 degrees and QPM is 30 degrees in triangle PQM. CPM =70 degrees so theta = 180- 60-70 Theta = 50 degrees
A very easy solution (it just took a few seconds to solve!) -
Draw a perpendicular from point Q. QM will be half of AB( because ABC and QMC are similar). And we know AB=QP. So, Triangle PQM is a 30-60-90 Triangle. Thus, we know that angle PQM is 60°. Angle MQC is 70°. So angle theta= 180-(60+70)=50°.
Hope this helps!
Very elegant solution!😊😊😊
@@phungpham1725 Thanks!
Let CQ = QA = a and AB = PQ = b.
Draw QT, where T is the point on BC where QT is perpendicular to BC. As QT is also parallel to AB, then ∠CQT and ∠CAB are corresponding angles and thus ∠CQT = ∠CAB = 90°-20° = 70°. As ∠QTC = ∠ABC = 90° and ∠C is common, then ∆QTC and ∆ABC are similar.
QT/AB = CQ/CA
QT/b = a/2a = 1/2
QT = b/2
Let ∠PQT = α.
cosα = QT/PQ = (b/2)/b
cosα = 1/2
α = cos⁻¹(1/2) = 60°
θ + α + 70° = 180°
θ + 60° + 70° = 180°
θ + 130° = 180°
[ θ = 180° - 130° = 50° ]
R es la proyección ortogonal de Q sobre BC---> Si AB=a---> QR=a/2---> sen(QPR)=(a/2)/a=1/2---> QPRº=30º ---> AQPº=20º+30º=50º.
Gracias y saludos.
Assign a value of 1 to line AC. At 20 degrees, line AB (sine of 20 degrees) will equal .3420. Line QP is thus .3420. Line QC is .5 . Going Pythagorean, consider triangle PQC. Angle at C is 20 degrees. Angle at P is 30 degrees. Angle at apex Q is 130 degrees. Line QM is .1710 . Indicated exterior angle at Q is 130-180 degrees or 50 degrees. Reasonably simple problem.
Nice solution. Congrats.
*_Uma solução bastante elegante!_* Parabéns!🎉🎉🎉
Tracing the line BQ and setting AQ=QC=x, AB=QP=y, we can write 2 identities with sines law:
On ABQ
Y/sin 40=x/sin 70
On PQC
Y/sin 20=x/sin alpha
Alpha is angle QPC
Comparing we get:
Sin alpha = sin 70*sin 20/sin 40
And knowing that
Sin 70 = sin (90-20)=Cos 20
Sin 40=2sin 20*cos 20
Sin alpha =1/2
Alpha =30
Theta = 20+30=50
t=2asin20...t/sin20=a/sin(θ-20)..asin20/sin(θ-20)=2asin20...1=2sin(θ-20)...θ-20=30..θ=50
AB = PQ = a, AQ = CQ = b, ∠CPQ = α
From △ABC we get:
tan(20°) = a/(2b)
Using law of sines in △CPQ we get
sin(α) / b = sin(20°) / a
sin(α) = b/a * sin(20°) = b/a * a/(2b) = 1/2
α = 30°
θ is an exterior angle of △CPQ, which is equal to the sum of its two opposite angles
θ = α + 20° = 30° + 20°
*θ = 50°*
Please correct it as :
From △ABC we get:
Sin(20°) = a/(2b)
These problems seem ten times easier after they have been worked out!
(I have deleted my false trails this time.)
Angle A =70 degrees
If QM is the perpendicular from Q to BC, it is (1/2) of a because triangles ABC and QMC are similar. PQM is 60 degrees and QPM is 30 degrees in triangle PQM.
CPM =70 degrees so theta = 180- 60-70
Theta = 50 degrees
I set AQ = 1, constructed BQ, found AQ equal to 2sin(20), then solved sin(theta - 20) = 1/2 from which theta = 50 degrees.
cos ( in ABMQ
We have sin(20)=AB/AC=PQ/2QC and PQ/sin(20)=QC/sin(θ-20) and from it 2sin(20)=sin(20)/sin(θ-20) so sin(θ-20)=1/2 so θ-20=30 and from it θ=50°
{20°A+20°C+90°B}={130°ACB ➖ 180°}=50°ACB 2^25 2^5^5 2^2^3^2^3 1^1^1^2^3 2^3 (ACB ➖ 3ACB+2).
Eu usei a Lei dos Senos. ;)
50 degree
asnwer=60 isit
asnwer=50 isit
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