A diagram shows a circle, centre O. AC is a diameter. angle BAC=31° D is a point on AC such that angle BDA is a right angle.?
Angle ABC = 90° since AC is a diameter and B is a point on the circumference of the circle. Angle BAC = 31° For the triangle ABC, Angle ABC + angle BAC + angle BCA = 180° Or, 90° + 31° + angle BCA = 180° Angle BCA = 180° - 90° - 31° = 90° - 31° = 59° For the triangle BDC, Angle BDC + angle BCD + angle DBC = 180° Or, 90° + 59 + angle DBC = 180° Or, Angle DBC = 180° - 90° - 59° = 90° - 59° = 31° Angle BOA = 2 × angle BCA = 2 × 59° = 118°
AB is the diameter and CD the chord which is equal to the radius and so half of AB. Let the centre be O. If you join C and D to O, and A to C and B to D, you will have three equilateral triangles each side of which is the same as the radius of the circle. The exterior angles CDE and DCE will be 60 degrees, and the third angle of the triangle EDC (which is the same as AEB) = 180 - 60 - 60 = 60 degrees.Proved.
The answer is angle 120°.Angle OPQ= angle OQP = 30° ie., Angle POQ= 120°. Also, angle PRQ= ½ reflex angle POQ.Source: NCERT
SukanyaSuppose the circle touches AC at Q (which is given as midpoint of AC)Compare the ∆s APQ and ABQ we have anglePAQ = angleBAQ (common angle) anglePQA = angleABQ in the alternate segment. Hence the two triangles are similar. Hence the corresponding sides are propl. Therefore we have AP/AQ = AQ/AB. Now AQ =(1/2)AC = (1/2)AB. So AQ/AB = 1/2Hence we have AP/(1/2)AB = 1/2So AP:AB = 1:4
Radius of CircumcircleGiven a triangle with vertices A, B and C, find a formula for the radius R of thecircumcircle .angle BOC is twice angle BAC since they intercept the same chord BC but BOC is a central angle Euclid's Elements, Book III, Proposition 20 and central and inscribed angles. In triangle BOC, BM = MC. Which leads to angle BOM = angle MOC.Which leads to saying that angle BOM = angle MOC = angle A. In the triangle BOM,sin(BOM) = BM / OBButBM =BC/2Which leads to:sin(BOM) = BC / (2*OB)angle BOM = angle A and OB is the radius R of the circle. Hencesin(A) = BC / (2R)or2R = BC / sin(A)and now using the sine law, we have2R = BC / sin(A) = AC / sin(B) = AB / sin(CIn this case, sides are=15 cm And angle=60°sin 60°=√3/2.So put the value to get R.
We need to use two geometry properties to answer this:(1) When two chords intersect inside a circle, the product of the two lengths on one chord = the product of the two lengths on the other chord.So AE * EB = CE * ED2 * 6 = 4 * ED,12 = 4EDED = 3.So now we know the lengths of all 4 of the little segments on the chords.(2) The center of a circle is on the perpendicular bisector of any chord.Let’s locate the chords conveniently on an x/y axis. We can put E at the origin, and the chords will fall along the x and y axes. Locate A and B on the x-axis and C and D on the y-axis withA = (-2, 0) and B = (6, 0)C = (0, 4) and D = (0, -3)Now you can easily find the midpoints of the two chord by averaging the endpoints (or just looking at your graph).AB: (2, 0) CD : (0, .5)Since the chords are vertical and horizontal, their intersection is easy to see. The intersection must be the center of the circle.Center = (2, .5)The length r of the radius would be the distance from the Center to any of the four points A, B, C, or D. We can use the Pythagorean theorem.Let’s use point C: (0, 4)r^2 = (2–0)^2 + (4 - .5)^2r^2 = 4 + 49/4 = ( 16 + 49)/ 4 = 65 / 4So r =( Sqrt65)/ 2The diameter is twice that: d = Sqrt 65You can confirm this by doing the same calculation with A, B, or D.
Draw the chord BD.By the Inscribed angle Theorem, [math]\angle BDC = \frac{1}{2}\angle BOC[/math][math]\angle ABD = \frac{1}{2}\angle AOD[/math]Consider the triangle BPD. By Exterior angle theorem,[math]\angle BPC = \angle ABD + \angle BDC[/math][math]= \frac{1}{2}(\angle AOD + \angle BOC)[/math]
CONSTRUCTION: Extend QS to ST , such that QS = STNow, PTRQ is a parallelogram ( as its diagonals bisect each other)< PQR = 90° ( given)=> PTRQ is a rectangle. ( //gm with an angle = 90° is rectangle)=> PR = QT ( diagonals of rectangles are equal)=> QS = PR/2=> QS = 10/5 = 5 cm
By Pythagoras theorem , ac is 10Now s is 12Area of triangle is 24Area is r *sThus, r = 24/12 = 2 cm