Advertisements
Advertisements
Question
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate : ∠BDC
Solution
Given that BD is a diameter of the circle.
The angle in a semicircle is a right angle.
∴ ∠ BCD = 90°
Also given that ∠ DBC = 58°
In Δ BDC ,
∠ BCD + ∠ BCD + ∠ BDC = 180°
⇒ 58° + 90° + ∠ BDC = 180°
⇒ 148° + ∠ BDC = 180°
⇒ ∠ BDC = 180° - 148°
⇒ ∠ BDC = 32°
APPEARS IN
RELATED QUESTIONS
In the given figure, SP is bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that : SQ = SR.
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:
- ∠ONL + ∠OML = 180°
- ∠BAM + ∠BMA
- ALOB is a cyclic quadrilateral.
Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
In the given figure, O is the centre of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find:
- ∠BOD
- ∠BPD
In a circle with centre O , chords AB and CD intersets inside the circle at E . Prove that ∠ AOC = ∠ BOD = 2 ∠ AEC.
In following figure , Δ PQR is an isosceles teiangle with PQ = PR and m ∠ PQR = 35° .Find m ∠ QSR and ∠ QTR
In the following figure, Prove that AD is parallel to FE.
In the adjoining figure, AB is the diameter of the circle with centre O. If ∠BCD = 120°, calculate:
(i) ∠BAD (ii) ∠DBA
In the figure , Δ PQR is an isosceles triangle with PQ = PR, and m ∠ PQR = 35°. Find m ∠ QSR and ∠ QTR.
In the given figure, the sides of the quadrilateral PQRS touches the circle at A, B, C and D. If RC = 4 cm, RQ = 7 cm and PD = 5 cm. Find the length of PQ:
