Advertisements
Advertisements
Question
In the below fig. O is the centre of the circle. Find ∠BAC.
Advertisements
Solution
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
RELATED QUESTIONS
Fill in the blank:
A circle divides the plane, on which it lies, in ............ parts.
In the given figure, O is the centre of the circle, prove that ∠x = ∠y + ∠z.
In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half the hypotenuse.
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, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
In the given figure, if O is the circumcentre of ∠ABC, then find the value of ∠OBC + ∠BAC.
In the following figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC = `1/2` (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).
