Mathematics
In the given diagram, O is the centre of the circle and the tangent DE touches the circle at B. If ∠ADB = 32°. Find the values of x and y.
Circles
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Answer
Given,
∠ADB = 32°
From figure,
∠ABC = 90° (Angle in a semicircle is a right angle)
We know that,
Angle between a tangent and a chord through point of contact is equal to an angle in the alternate segment.
∠BCA = ∠ABE = y
∠DBC = ∠BAC = x
Since, DE is a straight line, thus
⇒ ∠DBC + ∠ABC + ∠ABE = 180°
⇒ x + 90° + y = 180°
⇒ x + y = 180° - 90°
⇒ x + y = 90° ….(1)
In triangle ADB,
⇒ ∠ADB + ∠BAD + ∠ABD = 180°
⇒ 32° + x + (∠DBC + ∠ABC) = 180°
⇒ 32° + x + x + 90° = 180°
⇒ 2x + 122° = 180°
⇒ 2x = 180° - 122°
⇒ 2x = 58°
⇒ x = = 29°.
Substituting value of x in equation (1), we get :
⇒ 29° + y = 90°
⇒ y = 90° - 29°
⇒ y = 61°.
Hence, x = 29° and y = 61°.
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