In the given figure, O is the centre of the circle. ∠OABand ∠OCB are 30° and 40°
respectively. Find ∠AOC . Show your steps of working.
respectively. Find ∠AOC . Show your steps of working.
Exercise 17 (A) | Q 1 | Page 257
Download our app for free and get started
Join AC,
Let ∠OAC = ∠OCA = x (say)
∴ ∠AOC =180° -2x
Also, ∠BAC = 30°+ x
In ΔABC,
∠ABC =180°- ∠BAC -∠BCA
=180° - (30°+x) - (40°+ x ) = 110°- 2x
Now, ∠AOC = ∠2 ABC
(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°
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionIn the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: ∠ ACB.
Hence, show that AC is a diameter.
- 2View SolutionAB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find: ∠ BPR
- 3View SolutionIn the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate : ∠BDC
- 4View SolutionIn the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40°, find : ∠ADB
- 5View SolutionIn the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, Calculate ∠DEC .
- 6In the given figure, $A O B$ is a diameter and $D C$ is parallel to $A B$. If $\angle C A B=x^0$; find (in terms of $x$ ) the values of: $\angle$ ADC.View Solution
- 7View SolutionIn the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; Find:
(i) ∠ACB, (ii) ∠OBC, (iii) ∠OAB, (iv) ∠CBA.
- 8View SolutionIn the given figure, SP is bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR .
- 9View SolutionIn a regular pentagon ABCDE, Inscribed in a circle; find ratio between angle EDA and angle ADC.
- 10View SolutionIn the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.
Calculate : ∠ NRM