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प्रश्न
In the figure given alongside, AD is the diameter of the circle. If ∠ BCD = 130°, Calculate: (i) ∠ DAB (ii) ∠ ADB.
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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°
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In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
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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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Also, show that the ΔAOD is an equilateral triangle.
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Calculate:
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