Place point Q on half-line CB so that triangle QAC becomes a 30-75-75 isosceles triangle. Angle AQC = 30 degrees, line segment AQ = 2x = line segment QC At this time, the location of point Q is to the left or right of point B, or coincides with point B. 1) Q-B-C: line segment AQ = 2x = line segment QC > line segment BC (=2x), error 2) B-Q-C: line segment AQ = 2x = line segment QC < line segment BC (=2x), error 3) Therefore, point Q is located at point B and θ=30 degrees.
Tita =30 degree #=read as angle *=read as square root ^=read as to the power In the APC triangle, #ACP=75 tan 75=AP/PC (2+*3)=x/PC PC=X/(2+*3)=X(2-*3)/(4-3) PC=X(2-*3)=2X-*3X BP=BC-PC=2X-(2-*3X) =2X-2X+*3X=*3X Because in the ABP triangle #BPA=90 AB=*(AP^2+BP^2) AB=*{X^2+(*3X)^2} =*(X^2+3X^2)=*(4X^2) =2X SO, AB=2x, AP=x Sinb=AP/AB =X/2X=1/2 Sinb= sin 30degree b= 30 degree Thita = 30 degree
PC = xtan15= (2-rt3)x => BP = 2x-PC = rt3x => AB^2 = x^2 + 3x^2= 4x^2 => AB = 2x => ABC isoceles => angle BAC =75, => angle theta = 30 degrees. It is useful to be able to show tan 15 = 1/(2+rt3) = 2-rt3 without too much fuss.
Just make X = 1 and BC = 2. Going Pythagorean, liine AC =- .9659. Line PC becomes .249. That makes Line BP 1.75 (rounded). Tangent of angle at B = 1, and cosine = 1.75 Tan/Cos = .5714. Use arctan function on calculator , comes out to 30 degrees.
The 15°-75°-90° right triangle appear frequently enough in problems that it is a good idea to have its ratio of sides handy. ΔAPC is a 15°-75°-90° right triangle. The ratio of sides short:long:hypotenuse is (√3 - 1):(√3 + 1):2√2. Side PC is the short side and AP is the long side. So, PC is ((√3 - 1)/(√3 + 1)) times as long as AP, or has length ((√3 - 1)/(√3 + 1))x. We multiply the fraction by ((√3 - 1)/(√3 - 1)) and simplify to find PC has length (2 - √3)x or 2x - x√3. BP has length BC - PC = 2x - (2x - x√3) = x√3 So BP is √3 times as long as AP. The ratio of sides 30°-60°-90° right triangle, short:long:hypotenuse, is 1:√3:2. Here, we find that long side BP of ΔABP is √3 times as long as short side AP. This is the ratio for a 30°-60°-90° right triangle. So, AP is the short side, opposite the 30° angle and Θ = 30°, as Math Booster also found.
Extend CB. On CB call D such that B is midpoint of CD.We now have CD=4x.Triangle ACD with angle C equal 75° and base CD=4*height AP so ACD is right triangle with angle C=75° and angle ADC=15° and DC is hypotenuse with midpoint B so BA=BD then theta=angle ABC= 2* angle ADB=2*15°=30°
One could, alternatively, draw a perpendicular to BC at the center, let's call this midpoint "U" and then choose a point "V" in the perpendicular to BC through U and above BC in such a way that [angle VCB] = 15° and thus the triangles VCU and CAP are congruent (which implies that |CA| = |CV|) and [angle VCA] = 60°. As U is the midpoint of BC, |VB| = |VC| and, due to |VC| = |CA| and [angle VCA] = 60°, the triangle VCA is equilateral, which means that |VB| = |VC| = |VA|. This means that V is the circumcircle of triangle ABC and, as [angle CVA] = 60° due to VCA being equilateral, one must have that [angle CBA] = 30°. The motivation for this construction was simply the relation between the lengths of the height and the base of 1 to 2, the problem was screaming for similarity/congruency involving the base.
*Outra solução:* Lei dos senos no ∆ABC: AB/sen 75° = BC/ sen A. Note que: sen θ= x/AB e A=180 - (75° + θ), assim: AB=x/sen θ e sen A = 75° + θ. Daí, x/sen θ sen 75° = 2x/ sen (75° +θ) 1/sen θ sen 75° = 2/ sen (75° + θ) 2sen θ sen 75° = sen (75° + θ) _Fórmula trigonométrica:_ cos (A-B) - cos (A+B)=2sen A sen B 2sen θ sen 75°= cos (75°-θ) - cos (75°+θ). Daí cos (75°-θ) - cos (75°+θ)= sen (75°+θ) Ora, sen (75°+θ)= cos (90° -(75+θ), assim: sen (75°+θ)= cos (15°- θ). Assim, cos (75°- θ) - cos (75°+θ)= cos (15°- θ) cos (75°+θ) + cos (15°- θ) = cos (75°- θ) _Usando a fórmula trigonométrica:_ cos P + cos Q = 2cos(P+Q)/2 cos (P-Q)/2 Assim, cos (75°+θ) + cos (15°- θ) = 2cos 45° cos (30°+θ) Daí, 2cos 45° cos (30°+θ) = cos (75°- θ) cos (75°- θ)/cos (30°+θ)= cos 45°/cos 60° Logo 75°- θ = 45°→ *θ = 30°.* Nada fácil por trigonometria!
Идея: На продолжении отрезка BC берём точку L, такую , что AL=2x. Также на отрезке BC возьмем точку K, такую, что LK=2x. Угол ALP=30°. Углы LKA и LAK равны 75°. Треугольники ABC и ALK равны, по двум сторонам и углу между ними 75°. Тогда углы ALK и ABC равны, и равны 30°.
*Apontamento na solução:* Note que x=AP=a + a√3/2, isto é, x=a(2+√3)/2. Assim, BP=2x - a/2=2a(2+√3)/2 - a/2= a(3+2√3)/2. Assim, tag θ =AP/BP. Daí, tag θ =a(2+√3)/2 ÷a(3+2√3)/2 tag θ =(2+√3) ÷ (3+2√3) tag θ =√3(2+√3)/√3 ÷ (3+2√3) tag θ =(2√3+3)/√3 ÷ (3+2√3) tag θ = 1/√3 =√3/3, logo *θ = 30°.* Abraços!
Area of ABC=2xtimesx/2=x^2. Area of ABC= 2x (AC) sin 75degrees=2x(AC) cos15degrees=2x (AC))PC)/(AC). So, 2(PC)=x. Tan Theta=x/ (2x-x/2)=2/3=0.58. Thete=30degrees.
If I just try to simplify using the rules of sine and cosine we find that tan(θ)=2sin(75)/(√6+√2-2cos(75)) and since sin75=sin(30+45)=sin30*cos45+sin45*cos30=(√6+√2)/4 and likewise cos75=(√6-√2)/4 and by substitution we find tan(θ)=1/√3 so θ=30@@ritwikgupta3655
i didnt wath the full video but how does it takes 10 minutes? also the comments seems really complicated. its literally takes 10 seconds to solve it. just extend side BC 2x more to the left and connect it with point A. according to the rule (in the 15-75-90 triangle, 4x base is added to x height) you will have a 15-75-90 triangle. then use the rule of three (2x-2x-2x) and you can see 30 degrees.
what about extending side BC 2x more to the left and connect it with point A. according to the rule (in the 15-75-90 triangle, 4x base is added to x height) you will have a 15-75-90 triangle. then use the rule of three (2x-2x-2x) and you can see 30 degrees. it took 10 secs @@marioalb9726
Place point Q on half-line CB so that triangle QAC becomes a 30-75-75 isosceles triangle.
Angle AQC = 30 degrees, line segment AQ = 2x = line segment QC
At this time, the location of point Q is to the left or right of point B, or coincides with point B.
1) Q-B-C: line segment AQ = 2x = line segment QC > line segment BC (=2x), error
2) B-Q-C: line segment AQ = 2x = line segment QC < line segment BC (=2x), error
3) Therefore, point Q is located at point B and θ=30 degrees.
Tita =30 degree
#=read as angle
*=read as square root
^=read as to the power
In the APC triangle, #ACP=75
tan 75=AP/PC
(2+*3)=x/PC
PC=X/(2+*3)=X(2-*3)/(4-3)
PC=X(2-*3)=2X-*3X
BP=BC-PC=2X-(2-*3X)
=2X-2X+*3X=*3X
Because in the ABP triangle
#BPA=90
AB=*(AP^2+BP^2)
AB=*{X^2+(*3X)^2}
=*(X^2+3X^2)=*(4X^2)
=2X
SO,
AB=2x, AP=x
Sinb=AP/AB
=X/2X=1/2
Sinb= sin 30degree
b= 30 degree
Thita = 30 degree
PC = xtan15= (2-rt3)x => BP = 2x-PC = rt3x => AB^2 = x^2 + 3x^2= 4x^2 => AB = 2x => ABC isoceles => angle BAC =75, => angle theta = 30 degrees.
It is useful to be able to show tan 15 = 1/(2+rt3) = 2-rt3 without too much fuss.
Just make X = 1 and BC = 2. Going Pythagorean, liine AC =- .9659. Line PC becomes .249. That makes Line BP 1.75 (rounded). Tangent of angle at B = 1, and cosine = 1.75 Tan/Cos = .5714. Use arctan function on calculator , comes out to 30 degrees.
The 15°-75°-90° right triangle appear frequently enough in problems that it is a good idea to have its ratio of sides handy. ΔAPC is a 15°-75°-90° right triangle. The ratio of sides short:long:hypotenuse is (√3 - 1):(√3 + 1):2√2. Side PC is the short side and AP is the long side. So, PC is ((√3 - 1)/(√3 + 1)) times as long as AP, or has length ((√3 - 1)/(√3 + 1))x. We multiply the fraction by ((√3 - 1)/(√3 - 1)) and simplify to find PC has length (2 - √3)x or 2x - x√3. BP has length BC - PC = 2x - (2x - x√3) = x√3 So BP is √3 times as long as AP. The ratio of sides 30°-60°-90° right triangle, short:long:hypotenuse, is 1:√3:2. Here, we find that long side BP of ΔABP is √3 times as long as short side AP. This is the ratio for a 30°-60°-90° right triangle. So, AP is the short side, opposite the 30° angle and Θ = 30°, as Math Booster also found.
Extend CB. On CB call D such that B is midpoint of CD.We now have CD=4x.Triangle ACD with angle C equal 75° and base CD=4*height AP so ACD is right triangle with angle C=75° and angle ADC=15° and DC is hypotenuse with midpoint B so BA=BD then theta=angle ABC= 2* angle ADB=2*15°=30°
One could, alternatively, draw a perpendicular to BC at the center, let's call this midpoint "U" and then choose a point "V" in the perpendicular to BC through U and above BC in such a way that [angle VCB] = 15° and thus the triangles VCU and CAP are congruent (which implies that |CA| = |CV|) and [angle VCA] = 60°. As U is the midpoint of BC, |VB| = |VC| and, due to |VC| = |CA| and [angle VCA] = 60°, the triangle VCA is equilateral, which means that |VB| = |VC| = |VA|. This means that V is the circumcircle of triangle ABC and, as [angle CVA] = 60° due to VCA being equilateral, one must have that [angle CBA] = 30°. The motivation for this construction was simply the relation between the lengths of the height and the base of 1 to 2, the problem was screaming for similarity/congruency involving the base.
CotzT=2-cot75
Cot75=1-tan30/1+tan30=root3-1/root3+1=4-2root3/2=2-root3
CotT=root3
Theeta=30°
A=2x*x/2=(x/sinθ)(x/sin75)sin(90-θ+15)/2...2sinθsin75=cos(θ-15)..ctgθ=(2sin75-sin15)/cos15=2-tg15...θ=30
*Outra solução:*
Lei dos senos no ∆ABC:
AB/sen 75° = BC/ sen A. Note que:
sen θ= x/AB e A=180 - (75° + θ), assim:
AB=x/sen θ e sen A = 75° + θ. Daí,
x/sen θ sen 75° = 2x/ sen (75° +θ)
1/sen θ sen 75° = 2/ sen (75° + θ)
2sen θ sen 75° = sen (75° + θ)
_Fórmula trigonométrica:_
cos (A-B) - cos (A+B)=2sen A sen B
2sen θ sen 75°= cos (75°-θ) - cos (75°+θ). Daí
cos (75°-θ) - cos (75°+θ)= sen (75°+θ)
Ora, sen (75°+θ)= cos (90° -(75+θ), assim:
sen (75°+θ)= cos (15°- θ). Assim,
cos (75°- θ) - cos (75°+θ)= cos (15°- θ)
cos (75°+θ) + cos (15°- θ) = cos (75°- θ)
_Usando a fórmula trigonométrica:_
cos P + cos Q = 2cos(P+Q)/2 cos (P-Q)/2
Assim,
cos (75°+θ) + cos (15°- θ) = 2cos 45° cos (30°+θ)
Daí,
2cos 45° cos (30°+θ) = cos (75°- θ)
cos (75°- θ)/cos (30°+θ)= cos 45°/cos 60°
Logo 75°- θ = 45°→ *θ = 30°.*
Nada fácil por trigonometria!
Идея: На продолжении отрезка BC берём точку L, такую , что AL=2x.
Также на отрезке BC возьмем точку K, такую, что LK=2x. Угол ALP=30°. Углы LKA и LAK равны 75°. Треугольники ABC и ALK равны, по двум сторонам и углу между ними 75°. Тогда углы ALK и ABC равны, и равны 30°.
*Apontamento na solução:*
Note que x=AP=a + a√3/2, isto é, x=a(2+√3)/2. Assim,
BP=2x - a/2=2a(2+√3)/2 - a/2= a(3+2√3)/2. Assim,
tag θ =AP/BP. Daí,
tag θ =a(2+√3)/2 ÷a(3+2√3)/2
tag θ =(2+√3) ÷ (3+2√3)
tag θ =√3(2+√3)/√3 ÷ (3+2√3)
tag θ =(2√3+3)/√3 ÷ (3+2√3)
tag θ = 1/√3 =√3/3, logo
*θ = 30°.*
Abraços!
Area of ABC=2xtimesx/2=x^2. Area of ABC= 2x (AC) sin 75degrees=2x(AC) cos15degrees=2x (AC))PC)/(AC). So, 2(PC)=x.
Tan Theta=x/ (2x-x/2)=2/3=0.58. Thete=30degrees.
According to the rule of sines, we have AC/sin θ=2x/sin (θ+75) and AC=x/sin (75), and from this 2sin(θ)*sin (75)=sin (θ+75), so θ=30.
How? 2sin theta*sin75=sin(theta+75) is an identity. Does not solve for theta.
How? 2sin theta*sin75=sin(theta+75) is an identity. Does not solve for theta.
If I just try to simplify using the rules of sine and cosine we find that tan(θ)=2sin(75)/(√6+√2-2cos(75)) and since sin75=sin(30+45)=sin30*cos45+sin45*cos30=(√6+√2)/4 and likewise cos75=(√6-√2)/4 and by substitution we find tan(θ)=1/√3 so θ=30@@ritwikgupta3655
Tan 75degrees= 1/(2-root3). Tan75degrees = x/ |PC|=1/(2-root3) so |PC|= ( 2-root3)x. Therefore |BP|= 2x - [(2-root 3)x]= 2x - 2x + (root3)x=(root 3)x.Therefore tan theta= x/ (root3)x=1/ root3.So theta= 30 degrees.
Yes… simple, isn’t it?
{45°A+45°B+90°C}=180°ABC {2x+2x ➖ }=4x^2" 180°ABC/4x^2=40.20x^2 2^20.20x^2 2^10.2^10x^2 1^2^5.1^2^5x^1 1^1^2^1 2^1 (ABC ➖ 2ABC+1).
PC = y
tan75° = x/y
y = x/3,73....
tanB = x / (2x-y )
Bruch kürzen durch x:
tanB = 1 / ( 2 -y/x)
Einsetzen von y = x/3,73...
tanB = 1 / ( 2 - 1/3,73...)
tanB = 0,57735..
angle B= 30°
Obviously
x cot 75+x cot theta= 2x
2-√3 + cot theta= 2
cot theta = √3
theta= 30°
This construction is not simple to realise there has to be trigonometric solution.
Making an isosceles triangle is the most common construction.
@@MathBooster yeah, coincidently I had used it in a question I did today only
Sine rule:
(x/sinα)/sin75°= 2x/sin(α+75°)
sin(α+75°) = 2 sinα sin75°
sinα cos75°+cosα sin75°=2 sinα sin75°
cos75° + sin75°/tanα = 2 sin75°
sin75°/tanα = 2 sin75°- cos75°
1/tanα = 2 - 1/tan75°
tanα = 1 / (2-1/tan75°) = 1/√3
α = 30° ( Solved √ )
φ = 30°; ∆ ABC → ABC = θ = ?; BCA = 5φ/2; CAP = φ/2; BPA = 3φ; BC = BP + CP = (2x - k) + k
tan(φ/2) = k/x = 2 - √3 → k = x(2 - √3) → 2x - k = x√3 → AB = 2x → θ = φ
btw: sin(φ) = 1/2 → cos(φ) = √3/2 → sin(φ/2) = √((1/2)(1 - cos(φ))) = (√2/4)(√3 - 1) →
cos(φ/2) = (√2/4)(√3 + 1) → tan(φ/2) = sin(φ/2)/cos(φ/2) = 2 - √3 🙂
Nqse trekendeshi ABC eshte kende drejte ( nuk di gjuhen tuaj . Shoh vetem figuren) atehere kendi i kerkuar A eshte 25 grade.
i didnt wath the full video but how does it takes 10 minutes? also the comments seems really complicated. its literally takes 10 seconds to solve it. just extend side BC 2x more to the left and connect it with point A. according to the rule (in the 15-75-90 triangle, 4x base is added to x height) you will have a 15-75-90 triangle. then use the rule of three (2x-2x-2x) and you can see 30 degrees.
asnwer=30 isit
BP + PC = BC
x/tanθ + x/tan75° = 2x
1/tanθ + 1/tan75° = 2
1/tanθ = 2 - 1/tan75° = √3
tanθ = 1/√3 ---> θ=30° ( Solved √ )
Too complicated video solution !!!
I did this way too. It is the simplest solution
@ritwikgupta3655
Exacty !!! The simplest !!
1/tan75= tan 15
@@uferbenzo
BP + PC = BC
x/tanθ + x/tan75° = 2x
1/tanθ + 1/tan75° = 2
tan(90°-θ) = 2 - tan15° = √3
θ = 90°- atan(√3) = 30° ( Solved √ )
Almost the same !!!, ®uferberzo
what about extending side BC 2x more to the left and connect it with point A. according to the rule (in the 15-75-90 triangle, 4x base is added to x height) you will have a 15-75-90 triangle. then use the rule of three (2x-2x-2x) and you can see 30 degrees. it took 10 secs
@@marioalb9726
Digreets…
30degrees
Me falni kur pash me poshte kuptova se e kisha gabim. Ne trek AMC dybrinjenjeshem dalin kendet e bazes ......