Advertisements
Advertisements
प्रश्न
In Fig, Chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, Calculate ∠ DEC.
Advertisements
उत्तर
Consider the arc CDE. We find that ∠ CBE and ∠ CAE are the angles in the same segment of arc CDE.
∴ ∠ CAE = ∠ CBE
⇒ ∠ CAE = 65° ...( ∵ ∠ CBE = 65° )
Since AC is the diameter of the circle and the angle in a semicircle is a right angle.
Therefore, ∠ AEC = 90°.
Now, in Δ ACE, we have
∠ ACE + ∠ AEC + ∠ CAE = 180°
⇒ ∠ ACE + 90° + 65° = 180°
⇒ ∠ ACE = 25°
But ∠ DEC and ∠ ACE are alternate angles, because AC || DE.
∴ ∠ DEC = ∠ ACE = 25°.
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.
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.
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, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate : ∠BCD.
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.
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
In the given figure, AC is the diameter of the circle with center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED
