Advertisements
Advertisements
प्रश्न
In the below fig. O is the centre of the circle. Find ∠BAC.

Advertisements
उत्तर
WE have `/_AOB=80°`
and ∠AOC =110°
∴∠AOB+∠AOC+∠BOC=360°
⇒80°+110°+∠BOC=360°
⇒∠BOC=360°-80°-110°
⇒∠B0C=170°
By degree measure theorem
∠BOC=2∠BAC
⇒170°=2∠BAC
⇒∠BAC=`(170°)/2=85°`
APPEARS IN
संबंधित प्रश्न
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

In the given figure, two circles intersect at A and B. The centre of the smaller circle is Oand it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.

In the given figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.

In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is
If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.
In the following figure, ∠ACB = 40º. Find ∠OAB.

