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प्रश्न
In the given figure, BAD = 65°, ABD = 70°, BDC = 45°.
(i) Prove that AC is a diameter of the circle.
(ii) Find ACB.
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उत्तर
Given:
∠BAD = 65°
∠ABD = 70°
∠BDC = 45°
(i) In Δ ABD,
∠ BAD + ∠ABD + ∠ADB = 180°
65° + 70° + ∠ADB = 180° ....(Sum of three angles of a)
∠ADB = 180° - (65° + 70°)
∠ADB = 45°.
∠ ADC = ∠ADB + ∠BDC
45° + 45° = 90°
AC is the diameter of the circle. ....(Angle in a semi circle is 90°)
Proved.
(ii) ACB = ADB = 45° ....(Angles in the same segment of a circle)
संबंधित प्रश्न
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
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In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
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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 figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
