Mathematics
In the given figure, PAT is tangent at A and BD is a diameter of the circle. If ∠ABD = 28° and ∠BDC = 52°, find :
(i) ∠TAD
(ii) ∠BAD
(iii) ∠PAB
(iv) ∠CBD
Circles
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Answer
(i) ∠TAD = ∠ABD = 28° [Angles in the alternate segment are equal]
Hence, ∠TAD = 28°.
(ii) ∠BAD = 90° [Angle in the semicircle]
Hence, ∠BAD = 90°.
(iii) ∠PAB = ∠ADB [Angles in the alternate segment]
By angle sum property of triangle :
⇒ ∠ADB + ∠ABD + ∠BAD = 180°
⇒ ∠ADB + 28° + 90° = 180°
⇒ ∠ADB = 180° - (28° + 90°)
⇒ ∠ADB = 62°
⇒ ∠PAB = 62°.
Hence, ∠PAB = 62°.
(iv) In ΔBCD, since BD is the diameter ∠BCD = 90°.
By angle sum property of triangle,
⇒ ∠CBD + ∠BCD + ∠BDC = 180°
⇒ ∠CBD + 90° + 52° = 180°
⇒ ∠CBD = 180° - 90° - 52°
⇒ ∠CBD = 38°.
Hence, ∠CBD = 38°.
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