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प्रश्न
In figure, ABCD is a cyclic quadrilateral. <CBQ = 48° and x = 2y. Find the value of y.
योग
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उत्तर
Given:
ABCD is a cyclic quadrilateral.
angle ∠CBQ = 48° and x = 2y.
In triangle BCQ, ∠DCB (an exterior angle) = ∠CBQ + y = 48° + y.
In triangle APB, ∠ABP = ∠CBQ = 48° because they are vertically opposite angles.
Hence, angle ∠DAB = ∠ABP + x = 48° + x.
Since ABCD is cyclic, opposite angles sum to 180°: ∠DCB + ∠DAB = 180°
Substitute angles: (48° + y) + (48° + x) = 180°
Group terms: 48° + y + 48° + x = 180°
96° + x + y = 180°
Rearrange: x + y = 180° − 96°
= 84°
Given x = 2y, substitute: 2y + y = 84°
3y = 84°
y = `(84°)/3`
y = 28°
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