In the given figure AC is the diameter of the circle with centre O. Chord BD is perpendicular to AC. Express p in terms of x.

x / 2

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90^{\circ} + x / 2

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90^{\circ} - x / 2

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180^{\circ} - x

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Solution

The correct option is C $90^{\circ} - x / 2$
$∠ A O B = x \Rightarrow ∠ A D B = x / 2$
(Angle suntended by an arc at the centre is twice the angle at the remaining part of the $⨀$ ce of the circle )
Since $c h o r d B D ⊥ A C, ∠ A E D = 90^{\circ}$
$∴ I n △ A E D,$
$p = ∠ D A E = 180^{\circ} - (∠ A D E + ∠ A E D)$
$= 180^{\circ} - (x / 2 + 90^{\circ})$
$= 90^{\circ} - x / 2$

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In the figure AC is the diameter of circle, with centre O, chord BD is perpendicular to AC.

Write down the angles p, q, r in terms of .

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AC = CB,
$∠ x = 90^{\circ}$

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In the given figure AC is the diameter of the circle with centre O. Chord BD is perpendicular to AC. Express p in terms of x.

The correct option is C 90∘−x/2∠AOB=x⇒∠ADB=x/2(Angle suntended by an arc at the centre is twice the angle at the remaining part of the ⨀ce of the circle )Since chordBD⊥AC,∠AED=90∘∴In△AED,p=∠DAE=180∘−(∠ADE+∠AED)=180∘−(x/2+90∘)=90∘−x/2