Advertisements
Advertisements
प्रश्न
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
Advertisements
उत्तर
∠ADE = 95°(Given) Since,
OA = OB, so
∠OAB = ∠OBA
∠OAB = 30°
∠ADE + ∠ADC = 180°
(Linear pair)
95° + ∠ADC = 180°
∠ADC = 85°
We know that,
∠AOC = 2∠ADC
∠AOC = 2 × 85°
∠AOC = 170°
Since,
AO = OC
(Radii of circle)
∠OAC = ∠OCA
(Sides opposite to equal angle) ...(i)
In triangle OAC,
∠OAC + ∠OCA + ∠AOC = 180°
∠OAC + ∠OAC + 170° = 180°
[From (i)]
2∠OAC = 10°
∠OAC = 5°
Thus,
∠OAC = 5°
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
