Advertisements
Advertisements
प्रश्न
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC . Show your steps of working.
उत्तर
Join AC,
Let ∠OAC = ∠OCA = x ...(Say)
∴ ∠AOC =180° – 2x
Also, ∠BAC = 30° + x
∠BCA = 40° + x
In ΔABC,
∠ABC =180° – ∠BAC – ∠BCA
= 180° – (30° + x) – (40°+ x)
= 110° – 2x
Now, ∠AOC = ∠2ABC
(Angle at the centre is double the angle at the circumference subtended by the same chord)
`=>` 180° – 2x = 2(110° – 2x)
`=>` 2x = 40°
∴ x = 20°
∴ ∠AOC = 180° – 2 × 20° = 140°
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
- Prove that AC is a diameter of the circle.
- Find ∠ACB.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find : ∠PRB
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
- ∠ABC = 2∠APQ,
- ∠ACB = 2∠APR,
- `∠QPR = 90^circ - 1/2 ∠BAC`.
The given figure shows a circle with centre O and ∠ABP = 42°.
Calculate the measure of:
- ∠PQB
- ∠QPB + ∠PBQ
If two sides of a cyclic quadrilateral are parallel; prove that:
- its other two sides are equal.
- its diagonals are equal.
In the given figure, ∠ACE = 43° and ∠CAF = 62°; find the values of a, b and c.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in ∠ABC.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: ∠ ACB.
Hence, show that AC is a diameter.
If I is the incentre of triangle ABC and AI when produced meets the cicrumcircle of triangle ABC in points D. f ∠BAC = 66° and ∠ABC = 80°. Calculate : ∠BIC.
In the following figure, O is the centre of the circle, ∠ PBA = 42°.
Calculate:
(i) ∠ APB
(ii) ∠PQB
(iii) ∠ AQB
