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
Prove that the parallelogram, inscribed in a circle, is a rectangle.
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.

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.

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.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
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 following figure, AD is the diameter of the circle with centre O. chords AB, BC and CD are equal. If ∠DEF = 110°, Calculate: ∠FAB.

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 center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED

