<TQR= 62 deg. Proof: OQ is radius which is perpendicular to tangent PQR at point Q. Hence, <PQO = 90 deg. Now, <POQ = 180-34-90= 56 deg(sum of <s in a triangle) But <POQ= < SOT= 56 deg(vert. opp <s) Triangle SOT is Isosceles wit OS = OT Then <OST= <OTS(base angles) <OST= 180-56= 124/2= 62 deg. Finally, <OST= <TQR = 62 deg(<s on alt. segment)
Let's consider an equilateral triangle ABC with a side length equal to 'a'. If a perpendicular line is drawn from one of the vertices, to the side of the triangle, the perpendicular line divides the 60° angle at the vertex and the side of the triangle into two halves, that is 30° and a\2. the a/2 is the side opposite 30°, and the longest side is opposite the 90° formed by the perpendicular line and the side of the triangle remains a.
Much Simpler solution: 1) extend quarter circle from point A to point C on extension of line BO forming a semi circle of diameter 60. 2) Define theta as 1/2 of angle PBO. Theta= atan(15/30) 3) PBC forms a right triangle (from theorem for a triangle inscribed in a semi-circle with the diameter as the hypotenuse) 4) PB=60*cos(2*theta)=36 (angle PBC=2*theta)
Si el ángulo R=70º→ Ángulo central correspondiente =2*70=140º→ Los radios por X e Y son perpendiculares a las tangentes XP e YP→ Ángulo P =360º-140º-90º-90º=40º. Gracias y un saludo.
Respected professor Starting with intersecting chord theorem a*b=c*d and found the remaining chord length as 16. Then applied 4r^=a^+b^+c^+d^ to find radius of circle. Then executed the perpendicular bisector of the chord theorem to determine the half lengths of chords measuring 22 cms and 20 cms in order to ascertain the vertical distances from center to the chords and also joined the points O and M to form triangle DOM and triangle MOB. Calculated both vertical distances in the above tringles DOM and triangle MOB by applying Pythagorean theorem r^-(chord/2)^. The area obtained for these two tringles by (b*h)/2 were finally deducted from the area of the triangle DMB for the same result 50 unit^. With honour 🙏
Let <PBO=theta Applying cosine law to triangle PBO cos(theta)=(5^2+5^2-4^2)/(2*5*5)=17/25 sin(theta)=sqrt(1-cos(theta)^2)=sqrt(1-329/625)=4/25*sqrt(21) PD=5*sin(theta)=4/5*sqrt(21) PQ=2*PD=8/5*sqrt(21)
cos(x)=(10^2+10^2-12^2)/2(10)(10) So cos(x)=7/25 Cos(2x)=(R^2+R^2-144)/2R^2 Cos(2x)=(R^2-72)/R^2 2cos^2(x)-1=(R^2-72)/R^2 So R=25/4 So area of the circle=π(25/4)^2=625π/16=122.72.❤❤❤
Aclaraciones a mi comentario anterior: Si CA=6=2*3 y AB=8=2*4→ CB=2*5=10 → CA=6=CN→ CB=CN+NB=6+4→ Cómo MNB es semejante a CAB→ Si CN=4→MN=3=MA= Radio del semicírculo→ En el triángulo rectángulo MAC, la razón entre catetos es MA/AC=3/6=1/2→ El triángulo QPC es semejante a MAC→ Si QP=r→ PC=2r→ AP=6-2r→ Si "T" es la proyección ortogonal de Q sobre AM→ El triángulo rectángulo MTQ tiene lados MT=MA-QP=3-r ; TQ=AP=6-2r ; QM=3+r.
10=2*5 ; (12/2)=6=2*3 ---> h=2*4=8 ---> Potencia del punto medio de AB respecto a la circunferencia =6*6=8a---> a=9/2---> Radio =r=[8+(9/2)]/2 =25/4---> Área del círculo=πr² =625π/16 =122,71846...ud². Gracias y saludos.
Use the circumradius formula for triangles. Circumradius = side/divided by 2sine of opposite angle. CB is 10 cm in length. / 2sine of angle CAB ( 53.13) degrees , is 1.599 = circumradius works out to 6.2539. Using area of circle formula. Area is: 122.8720
Alternate approach: Rt triangle BOM has hypotenuse 5 and base 3. Therefore, OH = 4 = radius of circle. Now consider rt triangles PDO and PBD. Let DO = x and DB = (5 - x). From triangle PDO, (PD)^2 = (4)^2 - x^2. From triangle PBD, (PD)^2 = (5)^2 - (5 - x)^2. Equating these two expressions for (PD)^2 and solving for x gives x = 1.6. Knowing two sides of rt triangle PDO we can solve for (PD)^2 = 16 - (1.6)^2 = 13.44. This gives PD = 3.666 and since PQ = 2(PD) we have PQ = 7.332.
I got an alternative solution with trigonometry. Let C be on segment BP and OC is perpendicular to BP, then BP=2*BC …..(1) Let angle OBP=b tan(b/2)=(AO/2)/BO=15/30=1/2 -> cos(b/2)=2/sqrt(5) BC=BO*cos b=30*(2*cos(b/2)^2-1)=18 replacing to (1) BP=2*18=36
Yes, after sorting out that the radius is sqrt(125), I got DMB as (12*16)/2 = 96. Then subtracted the 2*5 rectangle, a (10*5)/2 triangle, and a (2*11)/2 triangle. 96 -(10+25+11) = 96 - 46 = 50 un^2
Yes. Draw two vertical lines from the circle centers to the 18 cm side. The distance between them is 18 - 5 - 8 = 5 cm. Form a right triangle with hypotenuse between circle centers. The height of the triangle is Sqrt( ((8 + 5) ^2) - 5^2) = sqrt(169 - 25) = sqrt(144) = 12. Then X = 12 + 5 + 8 = 25 and the Area of the rectangle is 18 * 25 = 20 * 25 - 50 = 450 cm ^2. I'll watch the video and see if we agree.
@@Maths_physicshubs Well, no. I just extended QT, then a perpendicular from SO to intersect that. You then have a right triangle hypotenuse OQ = 14, one leg = 5 + 2 = 7 and the other leg parallel and equal to ST. So Pythagoras gives you ST = 7√3 Looking into the transverse common tangent theorem, I can see how that would eventually lead you to the same place, but they still seem pretty superfluous to my mind.
*It's easy to show that this diagram is impossible. Since angle QRS (which is the same as angle PRS) is 40°, that means that angle QOS is 80° due to the inscribed angle theorem. The radius OS is perpendicular to the tangent PS, so the angle PSO would be 90°. But then the triangle PSO would have angles of 40°, 80° and 90°, which is impossible as the sum of the interior angles of a triangle **_must_** be 180° and 40+80+90=210°.*
∠QRP = 35° so by the inscribed angle theorem ∠QOP = 70° TQ is a tangent to the circle, so TQ⊥QO ∴ △TQO is right-angled with ∠TQO = 90° so by the properties of a △: ∠QTO =180 - (90 + 70) = 20°
Hi! Suggestion for a slightly more satisfying solution because it involves smaller numbers and immediate mental resolution (very subjective criteria, I know): QPR = 55 and TQP = 35, as you noted QPR is an external angle -> QPR = PTQ + PQT -> 55 = x+35
