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Empezando por el vértice inferior izquierdo y en sentido horario nombramos vértices exteriores de la figura: A, B, C, D y E. Llamamos H a la proyección ortogonal de B sobre AE. El ángulo en A es la suma de bº y aº---> tg(b+a)=(½+⅓)/[1-(½*⅓)]=1---> bº+aº=45º. Los triángulos CDA, CBD y DBA son semejantes y la razón entre los catetos respectivos es ½. La razón de semejanza entre los triángulos CBD y DBA es r=½. (2X)²=2²+6²---> X=√10=CD---> DA=2√10---> AC²=(√10)²+(2√10)²---> AC=5√2 ---> CB=½*½*BA---> CB=√2--->BA=4√2---> AH=4√2/√2=4---> HE=6-4=2---> Área EDB =2*HE/2=2 u². Gracias y un saludo cordial.

Pytagorean theorem: (2x)²=6²+2² --> x= √10m Internal angles: α=atan(1/3) ; β=atan(1/2) ; α+β=45° Height of shaded triangle: h= 6 - (2x)cosβ.cos(α+β) h= 6 - (2√10).(2/√5)/√2 = 6 - 4 = 2m Area of shaded triangle: A = ½b.h = ½2.2 = 2m² ( Solved √)

At 5:10, we have two triangles which share a hypotenuse of length 2√(10). (All linear dimensions in m.) One of them has a side of length 2√2 and an unknown side length. We use the Pythagorean theorem to find that the unknown side length is 4√2. We note that tan(a) = 2/6 = 1/3 and tan(b) = √(10)/√(20) = 1/2. We use the tangent sum of angles formula to find tan(a + b) = 1, or a + b = 45°. Drop a perpendicular to the line segment of length 6 from the 90° vertex of the right triangle with sides 2√2 and 4√2 and hypotenuse 2√(10). We have just formed an isosceles right triangle with hypotenuse 4√2, so the sides have length 4. If the side of the yellow triangle with length 2 is considered its base, its height is 6 - 4 = 2. The area is (1/2)bh = (1/2)(2)(2) = 2 m², as Math and Engineering also found.

Before looking at your demonstration : Yellow area = 1/2*2*2*sqrt(2)*sin(Pi/2-arcsin(1/sqrt(5))+arcsin(3/sqrt(10))) Yellow area = 2*sqrt(2)*cos(arcsin(1/sqrt(5))-arcsin(3/sqrt(10))) Yellow area = 2*sqrt(2)*(cos(arcsin(1/sqrt(5)))*cos(arcsin(3/sqrt(10)))+sin(arcsin(1/sqrt(5)))*sin(arcsin(3/sqrt(10)))) Yellow area = 2*sqrt(2)*(2/sqrt(5)*1/sqrt(10)+1/sqrt(5)*3/sqrt(10)) Yellow area = 2*sqrt(2)*(2/sqrt(50)+3/sqrt(50)) Yellow area = 2*sqrt(2)*5/sqrt(50) Yellow area = 2 m²

In another way, we can calculate the length of the third side of the yellow triangle using Ptolemy's theorem. Let this side be x, so it will be (2*4√2))+(2√2*6)=x*2√10, and from this x=2√5. We calculate the semi-perimeter of the triangle, where s=(2+2√2+2√5)/2=1+√2+√5, and from this the area of the triangle is equal to A=√(s*(s-2)*(s-2√2)*(s-2√5))=√4=2

@MathandEngineering : Thanks for this interesting video. (i found a demonstration similar (or worse) to yours a few days ago but it wasn't worth mentioning) @marioalb9726 indirectly put me on the trail of demonstrations that might be interesting. Like you 2 : * (PS) bissector of angle (TPV) then : y=4*d/h * (TS) bissector of angle (RTV) then : z=a*d/h (z=a*y/4) (angle (STV)=angle (PTV) - angle (RTS)=90°-45°=45°=angle(RTS)) * tan(n)=PQ/QR=TV/RT x/12=z/a then : x/12=d/h (and y=4*d/h=4*x/12=x/3) Theorem : " The interior angle bisectors of a triangle are concurrent in a point called the incenter of the triangle " Then : (VS) is the third bissector of triangle (PTV) We can verify it because : z/a=d/h Method 1 : * tan(angle(PVQ))=tan(angle(RVT)) (because (VS) bissector of angle (PVT) ; then PQV and RTV are similar triangles) PQ/QV=RT/TV x/(12+4+y)=a/z=12/x x/(16+x/3)=12/x x^2=(16+x/3)*12 x^2=192+4*x x^2-4*x-192=0 ( same equation that video at 11:30 ) x=16 Method 2 : (demonstration that PQS is an isosceles triangle) Let's call : angle (RPS)=angle(SPV)=q (VS) is bissector of angle (PVT) Then, angle(PVQ)=1/2*angle(PVT)=1/2*(90°-angle(TPV))=1/2*(90°-2*q)=45°-q angle(QSP)=180°-angle(PSV)=angle(SPV)+angle(PVQ)=q+(45°-q)=45° Then PQS is an isosceles right-angled triangle Then : PQ=QS x=PQ=QS=QR+RS=12+4=16 x=16 Following of Method 1 and 2 : @marioalb9726 "Secant Chords Theorem" ([PT] and [QV] secants in point R) or "cos(n)" implies that : 12/h=a/(4+y) a=12*(4+y)/h=12*(4+x/3)/h=(48+4*x)/h Area of triangle RTS : A=1/2*4*a*sin(n) A=2*a*(x/h) A=2*(48+4*x)/h*(x/h) A=(96*x+8*x^2)/h^2 A=(96*x+8*x^2)/(x^2+12^2) (Pythagoras : PR^2=PQ^2+QR^2) A=(96*16+8*16^2)/(16^2+12^2) A = 8,96 u²

At 8:00, we have a right triangle with a side of length 10 and hypotenuse 5√5, right angle vertex on the right side of the square. Applying the Pythagorean theorem, the remaining side (far right) has length 5. We also have a right triangle with one side being the bottom side of the square and hypotenuse 10. Applying the Pythagorean theorem, the remaining side (part of the right side of the square) has length 6. Drop a perpendicular from the farthest right vertex (3 line segments meet there) to the right side of the square. Two right triangles are formed, the top one having hypotenuse 5 (side of the right triangle with side of length 10 and hypotenuse 5√5). The right triangle with sides 6 and 8 and hypotenuse 10 is similar to the newly formed triangle, all dimensions twice that of corresponding sides. The vertical side of the newly formed triangle, which is a portion of the right side of the square, has length 4 (because its corresponding side on the similar triangle has length 8 and it is half as long), leaving 2 for the distance from the right angle vertex to the bottom right corner of the square. If the side with length 8 of the blue shaded triangle is considered the base, the height is 2 and the area is 8 m². (All linear dimensions in m, areas in m².)

L = 2m and z=2*60=m-60 L = 2(3*60) = 6*60 L = 360 mm (Solved √ )

Bisector theorem: h/d=a/z=4/y Similarity of triangles: a/z=12/x =4/y --> y=x/3 h²/d²=4²/y²=16/(x/3)² Pytagorean theorem, twice: (x²+12²)/(x²+(y+12+4)²)=(16*3²)/x² (x²+12²)/(x²+(x/3+16)²)=(16*9)/x² (x²+12²)/(x²+(x²/9+32/3 x+256))=144/x² (x²+12²)/(10/9 x²+32/3 x + 256)=144/x² x²(x²+144)=144.(10/9 x²+32/3 x + 256) x⁴+144x²=160 x²+1536x+36864 x⁴-16x²-1536x-36864=0 --> x=16cm (Fucking Isosceles triangle PQS, I couldn't see it, before) Intersecting chords theorem: a.h=12(4+y) =12(4+x/3)=48+4x Area of triangle: A= ½.4.a.sinβ <--- sinβ= x/h A= 2ax/h= 2xah/h²= 2x(48+4x)/h² A = (96x+8x²)/(x²+12²) A = 8,96 cm² ( Solved √ )

Las medianas de un triángulo se cortan en el baricentro "O" y éste las divide en dos tramos de ⅓ y ⅔ de la longitud respectiva ---> Si M y N son los puntos medios de AC y OC---> x=60 mm =OM/3---> OM=3*60=180---> OM=MB/3---> OB=2*180=360 mm. Gracias y un saludo cordial.

(4)^2(8)^2(12)^2={16+64144}=224ABCDEFG/180°=1.44ABCDEFG 1.2^2^2^2 1.1^1^1^2 1^2 (ABCDEFG ➖ 2ABCDEFG+1).

{60°A60°B+60°C}=180°ABC 60^60^60ABC 2^3^2^3^2^3ABC 1^1^1^1^2^3ABC 2^3ABC (ABC ➖ 3ABC+2).

The red triangle is 30-60-90 degrees so 4+y = 4√3

(8km)^2=64km^2 90/64km^2=1.26km 1.2^13 1.2^13^1 1.2^1^1 2^1 (km ➖ 2km+1).

a = 8/√2 = 4√2 km Base of rectangle: b = 8/2 = 4 km Height of rectangle: h.(h+2b)=a² (inters.chords theorem) h²+2hb = a² h² + 8h - 32 = 0 --> h=2,9282 km Area of rectangle : A = b.h = 11,713 km² ( Solved √)

(4 km )/ (8 km)=1/2=cos(60°) then : (4 km + y)/ (8 km)=sin(60°)=sqrt(3)/2 (4+y)/8=sqrt(3)/2 4+y=4*sqrt(3) y=4*(sqrt(3)-1) km x=4 km then A(rectangle)=x*y=16*(sqrt(3)-1) km²

Base del rectángulo =b=8/2=4 ---> a=4√2 ---> Altura del rectángulo =h=[√(r²-(r/2)²)] -4 =4(√3 -1) ---> Área del rectángulo =b*h=16(√3 -1). Gracias y un saludo cordial

The height of the right triangle is equal to the length of the rectangle, i.e. H = 4. According to the intersecting chords theorem, 4*12 = (4+y)², and from this y = 4√3-4. Therefore, the area of the rectangle is equal to 4*(4√3-4) = 16√3-16.

On peut établir plus rapidement la valeur de y par la trigonométrie.

(4m)^2,=16m^2 {90°A+90°B+90°C+90°D}=360°ABCD/16m^2=2.40m^2ABCD 2.2^20m^2 1.1^2^10m^2 1^2^5m^2 1^1^1m^2 1m^2 (ABCDm➖ 2ABCDm+1).

This is marioalb9726's method, done with the correct value of <TSV. Let <QSP = α and <TSV = ß. Then, for ΔPQS, <PQS + <QPS + α = 180°. However, ΔPQS is isosceles, <PQS = <QPS, and <QPS = (180° - α)/2 = 90° - α/2. Similarly, for ΔTSV, <TVS = 90° - ß/2. <QPS = <RPV, <TVS = <RVP and <RPV + <RVP + <PRV = 180°. So, 90° - α/2 + 90° - ß/2 + 60° = 180° and α + ß = 120°. Area ΔRPV = (1/2)(PS)(QS)sin(<QSP) = (1/2)(s)(s)sin(α) = 16. Area ΔTSV = (1/2)(TS)(SV)sin(<TSV) = (1/2)(s)(s)sin(ß) = 30. So, (30)(1/2)(s)(s)sin(α) = (16)(1/2)(s)(s)sin(ß)(s), which simplifies to (8/15)sin(ß) = sin(α) or sin(α) = (8/15)sin(ß). However, ß = 120° - α, so sin(α) = ((8/15)sin(120° - α). Apply the sine difference of angles formula, sin(a - b) = sin(a)cos(b) - cos(a)sin(b), to sin(120 - α): sin(120° - α) = sin(120°)cos(α) - cos(120°)sin(α). Substituting (√3)/2 for sin(120°) and -1/2 for cos(120°), we find sin(120° - α) = (√3)/2)cos(α) +(sin(α))/2. So sin(α) = (8/15)((√3)/2)cos(α) +(sin(α))/2), sin(α) = (4/15)((√3)cos(α) +sin(α)), (11/15)sin(α) = ((4√3)/15)cos(α), sin(α)/cos(α) = (4√3)/11. Since tan(Θ) = sin(Θ)/cos(Θ), tan(α) = (4√3)/11. At thins point, I turn to my scientific calculator. Calculations are done to its full precision and shown as rounded off here. So, α = 32.2° and ß = 120° - α = 87.8°. From (1/2)(s)(s)sin(α) = 16, s = √(32/sin(α)) = 7.75. <RPV = 90° - α/2 = 73.9°. <RVP = 90° - ß/2 = 46.1°. PV = 2s = 15.5. PV/sin(60°) = RV/sin(<RPV), so RV = 17.19. Area ΔPVR = (1/2)(PV)(RV)(sin(<RVP) = (1/2)(15.5)(17.19)(sin(46.1°) = 96, as Math and Engineering found at 16:00. (My calculation is to the precision of my calculator, but matches his exact answer.) Subtract the sum of the 2 given areas, 16 + 30 = 46, to get area QRST = 50 U²

{16°p+30°TV+60°R}={106°PTVR+74°S}=180°PTVRS 360°/180°PTVRS=2PTVRS (PTVRS ➖ 2PTVRS+2).

(98)^2=82504 (126)^2=5276 {82504+5276}=89780/180°=1759.6 100^100^10^70^59^1.6 1^3^4^1^1.2^3 1^2^2.1^3 1^2.1^3 2.3 (x ➖ 3x+2).

Let's assume that <TSV=θ and from it <PSQ=120-θ and ST=SV=SQ=SP=a we have [STV]=30 and [SPQ]=16 and from it we conclude a²sinθ=60, and a²sin(120-θ)=32 Dividing the two equations we find 8sinθ=15sin(120-θ) this last equation is equivalent to tanθ=15√3. From this cosθ=1/26 and sinθ=(15√3)/26 so a²=60/sin θ=(104√3)/3 and since the triangle QST is equilateral its area is [QST]=(a²√3)/4=26 so the area of the quadrilateral QPVT is [QPVT]=30+16+26=72 The triangles RQT and RPV are similar and their similarity ratio is QT/PV=1/2 so if we assume that [RQT]=x then [RQT]/[RPV]=(x)/((x+72)=1/4 and solving this equation we find x=24 so the area of the quadrilateral QRST is [QRST]=26+24=50

Just wondering, why the host expands on basic aritmetic so much? Kind of annoying, audience is geometry lovers, knows aritmetic by default🤔? Problems itself are excelent, narrative not so much, sorry!

Original and impressive ! Here are my methods : P,Q,T,V are on the (semi)circle of center S and radius R. angle(RPV)=c ; angle(RVP)=d c+60°+d=180° then d=120°-c area(PQS)=16 PS*cos(c)*PS*sin(c)=16 1/2*R^2*sin(2*c)=16 R^2*sin(2*c)=32 area(VTS)=30 SV*cos(d)*SV*sin(d)=30 1/2*R^2*sin(2*d)=30 R^2*sin(2*d)=60 R^2*sin(240°-2*c)=60 R^2*(sin(240°)*cos(2*c)-cos(240°)*sin(2*c))=60 R^2*(-sqrt(3)/2*cos(2*c)-(-1/2)*sin(2*c))=60 R^2*(-sqrt(3)*cos(2*c)+sin(2*c))=2*60 -sqrt(3)*R^2*cos(2*c)+R^2*sin(2*c)=120 -sqrt(3)*R^2*cos(2*c)+32=120 sqrt(3)*R^2*cos(2*c)=-88 R^2*cos(2*c)=-88/sqrt(3) (R^2*cos(2*c))^2+(R^2*sin(2*c))^2=(-88/sqrt(3))^2+32^2 R^4*((cos(2*c))^2+(sin(2*c))^2)=88^2/3+32^2 R^4=8^2*(121/3+16) R^4=8^2*169/3 R^4=(8*13)^2/3 R^2=104/sqrt(3) Method 1 : we will notice that QST is equilateral and RQT similar to PRV with ratio=1/2 angle(QST)=180°-angle(PSQ)-angle(VST) angle(PSQ)=180°-2*angle(RPV)=180°-2*c angle(VST)=180°-2*angle(RVP)=180°-2*d angle(QST)=180°-(180°-2*c)-(180°-2*d) angle(QST)=2*(c+d)-180° angle(QST)=2*120°-180° angle(QST)=60° But QST is an isosceles triangle then : angle(SQT)=angle(STQ)=(180°-angle(QST))/2=(180°-60°)/2=60° Then, QST is an equilateral triangle. QT=QS=ST=R angle(RQT)=180°-angle(SQT)-angle(PQS) angle(RQT)=180°-60°-c angle(RQT)=120°-c angle(RQT)=d angle(RTQ)=180°-angle(STQ)-angle(STV) angle(RQT)=180°-60°-d angle(RQT)=120°-d angle(RQT)=c Triangle RQT is similar to triangle PRV : same angles : c, d, 60° QT=R=PV/2. Then : QR=RV/2 and RT=PR/2 and : Area(RQT)=(1/2)^2*Area(PRV) Area(RQT)=1/4*Area(PRV) 4*Area(RQT)=Area(PRV) 4*Area(RQT)=Area(RQT)+Area(QST)+Area(PQS)+Area(STV) 3*Area(RQT)=Area(QST)+Area(PQS)+Area(STV) Area(RQT)=1/3*Area(QST)+1/3*(Area(PQS)+Area(STV)) Area(RQT)=1/3*Area(QST)+1/3*(16+30) Area(QRTS)=Area(RQT)+Area(QST) Area(QRTS)=4/3*Area(QST)+1/3*(16+30) Area(QRTS)=4/3*sqrt(3)/4*R^2+46/3 Area(QRTS)=1/sqrt(3)*104/sqrt(3)+46/3 Area(QRTS)=104/3+46/3 Area(QRTS)=50 u² Method 2 : we will calculate tan(c) and tan(d) tan(2*c)=(R^2*sin(2*c))/(R^2*cos(2*c)) tan(2*c)=32/(-88/sqrt(3)) tan(2*c)=-4*sqrt(3)/11 With formula : tan(2*c)=2*tan(c)/(1-(tan(c))^2), we obtain the following second degree equation : 2*sqrt(3)*T^2-11*T-2*sqrt(3)=0 (with T=tan(c)) T=tan(c)=2*sqrt(3) (because we can see that tan(c)>0 then the other solution T=tan(c)=-sqrt(3)/6 is impossible) tan(d)=tan(120°-c) tan(d)=(tan(120°)-tan(c))/(1+tan(120°)*tan(c)) tan(d)=(-sqrt(3)-2*sqrt(3))/(1-sqrt(3)*2*sqrt(3)) tan(d)=-3*sqrt(3)/(-5) tan(d)=3*sqrt(3)/5 Area(PRV)=1/2*Base*Height Base=PV=2*R Height=Base*tan(c)*tan(d)/(tan(c)+tan(d)) : i don't explain this formula i know Area(PRV)=2*R^2*tan(c)*tan(d)/(tan(c)+tan(d)) Area(PRV)=2*104/sqrt(3)*2*sqrt(3)*3*sqrt(3)/5/(2*sqrt(3)+3*sqrt(3)/5) Area(PRV)=96 Area(PRV)=Area(QRTS)+Area(PQS)+Area(STV) 96=Area(QRTS)+16+30 Area(QRTS)=50 u²

Supposing that isosceles triangle TSV is a right triangle: Side of isosceles right triangle: ½s² = 30cm² ---> s= √60=2√15cm Base of larger triangle: b = 2s = 4√15 cm Angle RPV: α = 180°-60°-45°= 75° Sine rule: c/sin75°= b/sin60° --> c=17,279 cm Area of triangle RPV: A₁= ½.b.c.sin45° = 94,641 cm² Quadrilateral shaded area: A = A₁-30-16 = 48,64 cm² (which is an approximate result)

Supposing that isosceles triangle TSV is a right triangle: Side of Isosceles right triangle: ½s² = 30cm² ---> s= √60cm Internal angles: α = 180°-60°-45°= 75° (∠RPV) β = 180°-(2*75°)= 30° (∠QSP) δ = 90°-β = 60° , so ΔQST is equilateral γ = 180°-α-δ = 45° (∠RQT) Sine rule, triangle QRT: a/sin75°=s/sin60° --> a=8,6395 cm Triangle RQT: A₁= ½.a.s.sin45° = 23,66cm² Equilateral triangle QST: A₂ = √3/4 s² = 25,98 cm² Quadrilateral shaded area: A = A₁+A₂ = 49,64 cm² ( Which is an approximate result )

(6m)^2 (33m)^2={36m^2+1089m^2}=1125m^4 {60°A+60°B+60°C}=180°ABC 1125m^4/180°ABC=10013.1 10^1013.1 2^5^2^53. 1^1^2^13. 23. (ABC ➖ 3ABC+2).

{2a+2a ➖ }{15°A+45°B+90°C}=4a^2{150°ABC+30°D}=180°ABCD/4a^2=40a^4.20ABCD 2^20a^2^2.2^10 1^10a^1^1.2^2^5 2^5.1^1^1 2^1.1 2.1 (ABCD a ➖ 2ABCD a+1).

Solving the equation found at 3:40 using a half interval search: Let the linear dimensions be in m and areas in m². Then the equation becomes √(a² - 36) = √(a² - 1089) + √(a² - 1521). Let Δ = √(a² - 36) - (√(a² - 1089) + √(a² - 1521)), the difference between the two sides of the equation. We wish to find a value of a that will make that difference Δ as small as possible. We note that a cannot be less than 39 or a² - 1521 would be negative, producing an imaginary number for √(a² - 1521). For a = 39, Δ is positive. Experimentally, I find Δ is negative for a = 55. So, I write a computer program to perform a half interval search for a value of a which minimizes Δ, using search range limits of a = 39 (low end) and a = 55 (high end). I compute Δ for the midpoint, 47. If a negative result, I check 43 (half way between 39 and 47) otherwise 51 (half way between 47 and 55). I continue to zoom in by cutting the distance in half each time. My computer result is 42.000000000000 to within the precision of its calculations. Now, is 42 exact? I substitute a = 42 into the original equation and get √(42² - 36) = √(42² = 1089) + √(42² - 1521), which simplifies to √(1728) = √(675) + √(243). I square both sides: 1728 = 675 + 2(√(675))(√(243)) + 243, which simplifies to 1728 = 918 + 2(√(164025)), 1728 = 918 + 2(405), and 1728 = 1728. So, a = 42 is an exact answer. Here is my computer program, written in python, and its result: import math a = 47.0 da = 4.0 # define function to compute delta = ∆ def compute_delta(a): delta = math.sqrt(a*a - 36) - (math.sqrt(a*a - 1089) + math.sqrt(a*a - 1521)) return(delta) for n in range(0,51): # after 50 divisions by 2, da is sufficiently small delta = compute_delta(a) if (delta > 0): a = a + da else: a = a - da da = da/2 a = round(a , 12) # round a to 9 decimal places formatted = "{:.12f}".format(a) print(f"a = ",formatted) Output: a = 42.000000000000 === Code Execution Successful ===

We call the triangle ECD and let H be the perpendicular projection of point E on BC. We put <HEC=α and from it we get the equation 33tan(α)+39tan(60-α)=6tan(120-α) and this equation is equivalent to tan(α)=(45√3)/99 and from it cos(α)=√(1/1+tan²(α))=11/14 and from it a=33/cosα=33*14/11=42 and from it the area of triangle EBD is equal to (42²*√3)/4=441√3

PQT = 35° → tan⁡(35°) ≈ 7/10 → a = 70 = QS = TS QTR = α → tan⁡(α) ≈ 9/10 → area ∆RST = area ∆RQT - area ∆RQS ≈ 49(90)

(10)^2 (3m)^2={100+9m^2}=109m^2 {10°A+10°C+130°B}={150°ACB+30°D}=180°ACBD/109m^2=1.7..8 1.7.2^3 1.3^4.1^1^1 3^2^2.1 3^1^2.1 3^2 (ACBD ➖ 3ACBD+2).

φ = 30° → sin⁡(3φ) = 1; ∎ADNB → AB = AM + BM = 6 + 33 = DN BN = BC + CN = k + (y - k) = AD = y; MDC = DCM = CMD = 2φ → CM = CM = CD = a; ADM = α; CDN = β → α + β = φ → BMC = φ + α → cos⁡(φ + α) = cos⁡(φ)cos⁡(α) - sin⁡(φ)sin⁡(α) = (1/2a)(y√3 - 6) = 33/a → y = 24√3 → a^2 = 1764 → area ∆ CMD = a^2(√3/4) = 441√3

Let CDE be the equilateral triangle and move AB horizontally to the right until E be on line segment AB. Let <BEC= x and AED=y then x+y+60=180 ->x=120-y BE=EC*cos x ->33=a*cos(120-y) …….(1) AE=ED*cos y->6=a*cos y ……..(2) Dividing (1) by (2) 11/2=cos(120-y)/cos y =(cos 120*cos y+sin 120*sin y)/cos y=-1/2+sqrt(3)/2*tan y tan y=4*sqrt(3) from which sec y=sqrt(tan(y)^2+1)=7 ->cos y=1/7 replacing to (2) 6=a*1/7 ->a=42 Area CÈDE=sqrt(3)/4*a^2=441*sqrt(3)

α+β= 60° a = 33/sinα ; a = 6/sinβ Equalling: 33 sinβ = 6 sinα sin(60°-α) = 6/33 sinα ½(√3cosα - sinα) = 6/33 sinα √3/tanα - 1 = 12/33 √3/tanα = 45/33 tanα= 33√3/45 --> α=51,787° Triangle side: a = 33/sinα = 42 m Area of equilateral triangle: A = √3/4 a² = 441√3 m² A = 763,83 m² ( Solved √ )

I solved the problem using the sine rule on the full triangle and the triangle on the right. BCA = BAC = 10 --> ABC = 180-2*10 = 160 --> CBD = 20. Then BC / sin CBD = BD / sin BCD --> CD / sin (20) = BD / sin (100) = BD / sin (80) --> BD = CD * sin (80) / sin (20) = BC * cos (10) / sin (20). Now CD / sin (10) = 10√3 / sin ADC = 10√3 / sin (60). [Proof: ACB = 10 -> ACD = 110 -> ADC = 180 - (10+110) = 60.) --> CD = 10√3 * sin (10) / sin (60) --> BD = 10√3 * sin (10) * cos (10) / sin (20) / sin (60) = 5√3 * sin (20) / sin (20) / sin (60) = 5√3 / ((√3)/2) = 5*2 = 10m

QPT = 90-QTP = 90-35 = 55. Now QT/ sin (QPT) = PQ / sin (35) --> QT = 98 * sin (55) / sin (35) = 139.96 = 140. Now ST = QT - QS = QT/2 = 140/2 = 70. Then [RST] = QR*ST/2 = 126 * 70/2 = 63*70 = 4410 sq. units.

I obtained [ABD] = 32 mm EXACTLY. (Just as in the video) AD = √(12^2 + 4^2) = √160 = 4√10. Let midpoint of AE be called F. Then DF = √(6^2 + 4^2) = √52 = 2√13. Then [ABD] = AD*BD/2 * sin (ADB). Here BD = DF√2 = 2√26. Now, tan (ADE) = Y = 12/4 = 3 and tan (FDE) = X = 6/4 = 3/2. Then tan (ADF) = (Y-X) / (1+YX) = (3-3/2) / (1+3*3/2) = (3/2) / 1+9/2) = (3/2) / (11/2) = 3/11. Now, BDF=45 --> tan (BDF) = W = 1 and tan (ADF) = Z = 3/11. Then tan (ADB) = (W-Z) / (1+WZ) = (1-3/11) / (1+1*3/11) = (8/11) / (14/11) = 4/7. Thus sin (ADB) = 4 / √(4^2 + 7^2) = 4 / √(16+49) = 4/√65. Then [ABD] = (4√10 * 2√26)/2 * 4/√65 = (4*2*4/2) * √ (260/65) = (32/2) * √4 = 16*2 = 32.

(15)^2 (12)^2={225+144}=569/360°QPSR=1.209QPSR 1.10^20^9 1^2^3^2 1^2^3^1 2^3 (QPSR ➖ 3QPSR+2).

Thanks a lot professor for your efforts. I follow you from Algeria

BD=(sin100)*(sin10*)10√3/(sin60)(sin160) m

I solved with trigonometry and sines law in this way: On triangle BCD BC/ sin 60 = BD/sin 100 BC = BD*sin 60 /sin 100 (1) On triangle ABC AC/ sin 160 = BC/sin 10 BC = AC*sin 10/ sin 160 (2) Knowing that: sin 100 = sin (90 + 10) = sin(90 - 10) = cos 10 and sin 160 = sin(180 - 20) = sin 20 we can campare (1) with (2) (BD*√ 3/2)/cos 10 = 10√ 3*sin 10/sin 20 BD/2*sin 20 = 10*sin 10*cos 10 BD/2*sin 20 = 5*sin 20 BD = 10

Angle CDB= α α = 180°-100°-(2*10°) = 60° h = 5√3 sinα = 5√3.√3/2 = 7,5 mts. Similarly of right triangles: x/5√3 = 5√3 /7,5 x = 10 mts (Solved √ ) If the figure were drawn into scale, this will be seen clearly !!!

<ACB=<CAB=10 hence sin(<ABC)=sin(180-10-10)=sin(180-20)=sin 20 <ADC=180-10-10-100=60 sin(<ADC)=sin 60 Applying sine law to triangle ABC sin(<CAB)/BC=sin(<ABC)/AC ->sin 10/BC=sin 20/(10*sqrt(3) -> sin 10/BC=2*sin 10*cos 10/(10*sqrt(3) ->1/BC=cos 10/(5*sqrt(3)) …..(1) Similarly to triangle BCD with <BDC=<ADC sin(<BDC)/BC=sin(<BCD)/x ->sin 60/BC=sin 100/x -> sqrt(3)/(2*BC)=cos 10/x ….(2) Dividing (1) by (2) 2/sqrt(3)=x/(5*sqrt(3)) x=10

Let AB=BC=x, from which x/(sin10)=(10√3)/(sin(160)), then x=(10√3*sin (10))/sin(160), and we have BD/sin(100)=x/(sin (60))=(2x)/√3, then BD=(10√3*sin 10/sin 160)*(2sin 100)/√3=(20sin 10*sin100)/(2sin 80*cos80)=(10sin 10*sin 100)/(si100*sin 10)=10

Thankyou sir

Using Cachi's theorem for triangles ABC and BEG we find cos(a)=(AB²+BC²-AC²)/(2*AB*BC)=(BE²+BG²-EG²)/(2*BE*BG) then let AB=x. Then BC²=x²+16, BE²=x²+144, BG²=x²+576, so we get an equation with one unknown x. After simplification, we get an equivalent equation x⁴-72x²-10368=0, i.e. (x²+72)(x²-144)=0. Therefore, x=12, and from this, BG=12√5. Since the triangles ABG and CFG are similar and their similarity ratio is equal to CG,/BG=5/(3√5), the area of triangle CFG is (12*24/2)*(5/3√5)²=80.

I don't know why these approximate values at the beginning of the solution and mathematics is called an exact science. The final result for the area of the triangle RST was supposed to be A=3087/tan(35°) and tan(35°) remains like any irrational number √2, π, e... and if we want to write A≈4410 and it is preferable to mention even the accuracy of the approximation so that the approximate values do not differ from one person to another and thank you for the efforts.