Mathematics
In Fig. 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.
Answer
From figure,
BD is a straight line
⇒ ∠BEC + ∠DEC = 180° (Linear pairs)
⇒ 130° + ∠DEC = 180°
⇒ ∠DEC = 180° - 130°
⇒ ∠DEC = 50°.
In ∆ DEC,
⇒ ∠DEC + ∠EDC + ∠ECD = 180° (Angle sum property of a triangle)
⇒ 50° + ∠EDC + 20° = 180°
⇒ ∠EDC + 70° = 180°
⇒ ∠EDC = 180° - 70°
⇒ ∠EDC = 110°.
From figure,
⇒ ∠BDC = ∠EDC = 110°
We know that,
Angles in the same segment of a circle are equal.
⇒ ∠BAC = ∠BDC = 110°.
Hence, ∠BAC = 110°.
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.
In Fig. ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
In Fig. ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
