Advertisements
Advertisements
प्रश्न
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
Advertisements
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°
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
Prove that the rhombus, inscribed in a circle, is a square.
In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM
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.
In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
- ∠DAB,
- ∠DBA,
- ∠DBC,
- ∠ADC.
Also, show that the ΔAOD is an equilateral triangle.
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.
Using ruler and a compass only construct a semi-circle with diameter BC = 7cm. Locate a point A on the circumference of the semicircle such that A is equidistant from B and C. Complete the cyclic quadrilateral ABCD, such that D is equidistant from AB and BC. Measure ∠ADC and write it down.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBC
Also, show that the ΔAOD is an equilateral triangle.
