Advertisements
Advertisements
प्रश्न
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
पर्याय
130°
115°
65°
165°
उत्तर
115°
We have the following information in the following figure. Take a point P on the circle in the given figure and join AP and CP.
Since, the angle subtended by a chord at the centre is twice that of subtended atany part of the circle.
So, `angleAPC = (angleAOC)/2 = 130/2 = 65°`
Since `squareAPBP` is a cyclic quadrilateral and we known that opposite angles are supplementary.
Therefore,
\[ \Rightarrow \angle ABC + 65° = 180° \]
\[ \Rightarrow \angle ABC = 180° - 65° \]
\[ \Rightarrow \angle ABC = 115°\]
APPEARS IN
संबंधित प्रश्न
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
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.
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.