Advertisements
Advertisements
प्रश्न
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
उत्तर
(i) Since ABCD is a cyclic quadrilateral.
∴ Its Opposite angles are supplementary.
∴ ∠ DAB + ∠ BCD = 180°
⇒ ∠ DAB = 180° - ∠ BCD
⇒ ∠ DAB = 180° - 130°
⇒ ∠ DAB = 50°
(ii) Since, angle in the semicircle is a right angle.
∴ In Δ ABD, ∠ABD = 90°
Since, the sum of the angle of a triangle is 180°
∴ ∠ABD + ∠ADB + ∠ DAB = 180°
∴ 90° + ∠ADB + 50° = 180°
∠ADB = 180° - (90° + 50°)
∠ADB = 180° - 140°
∠ADB = 40°
APPEARS IN
संबंधित प्रश्न
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate :
- ∠DAB,
- ∠BDC.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.
In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠NRM
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠ADC
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
In the figure, ∠DBC = 58°. BD is diameter of the circle.
Calculate:
- ∠BDC
- ∠BEC
- ∠BAC
In the given figure AC is the diameter of the circle with centre O. CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°.
Calculate
- ∠BEC
- ∠BCD
- ∠CED