Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
Options
80º
50º
40º
30º
Advertisements
Solution
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to 50º.
Explanation:
Given, ABCD is a cyclic quadrilateral and ∠ADC = 140°.
We know that, sum of the opposite angles in a cyclic quadrilateral is 180°.
∠ADC + ∠ABC = 180°
⇒ 140° + ∠ABC = 180°
⇒ ∠ABC = 180° – 140°
∴ ∠ABC = 40°
Since, ∠ACB is an angle in a semi-circle.
∴ ∠ACB = 90°
In ΔABC, ∠BAC + ∠ACB + ∠ABC = 180° ...[By angle sum property of a triangle]
⇒ ∠BAC + 90° + 40° = 180°
⇒ ∠BAC = 180° – 130° = 50°
APPEARS IN
RELATED QUESTIONS
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
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 such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
