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प्रश्न
In the figure, circle with centre D touches the sides of ∠ACB at A. If ∠ACB = 52°, complete the activity to find the measure of ∠ADB.

Activity:
ln `square` ABCD,
∠CAD = ∠CBD = `square`° ....Tangent theorem
∴ ∠ACB + ∠CAD + ∠CBD + ∠ADB = `square`°
∴ 52° + 90° + 90° + ∠ADB = 360°
∴ ∠ADB + `square`° = 360°
∴ ∠ADB = 360° − 232°
∴ ∠ADB = `square`°
कृती
बेरीज
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उत्तर
ln `square` ABCD,
∠CAD = ∠CBD = \[\boxed{90}\]° ....Tangent theorem
∴ ∠ACB + ∠CAD + ∠CBD + ∠ADB = \[\boxed{360}\]°
∴ 52° + 90° + 90° + ∠ADB = 360°
∴ ∠ADB + \[\boxed{232}\] = 360°
∴ ∠ADB = 360° − 232°
∴ ∠ADB = \[\boxed{128}\]°
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