In the given figure, $A B$ is the diameter of a circle $C (O, r)$ . chord $C D$ is equal to radius $O C$ . If $A C a n d B D$ intersect at $P$ , find $∠ A P B$ .

30^{\circ}

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45^{\circ}

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60^{\circ}

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90^{\circ}

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Solution

The correct option is C $60^{\circ}$
$I n △ O C D O C = O B = C D ∴ ∠ C O D = 60 (A n g l e s o f e q u i l a t e r a l t r i a n g l e) A S ∠ C A D = \frac{1}{2} ∠ C O D = \frac{1}{2} \times 60 = 30 N o w ∠ A D P = ∠ A D B = 90 (A n g l e i n s e m i c i r c l e) I n △ A D P ∠ A D P + ∠ D P A + ∠ D A P = 180 \Rightarrow 90 + ∠ D P A + 30 = 180 \Rightarrow ∠ D P A = 180 - 120 = 60 = ∠ A P B H e n c e o p t i o n (c) i s r i g h t a n s w e r$

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In the given figure, AB is the diameter of a circle C(O,r). chord CD is equal to radius OC. If ACandBD intersect at P, find ∠APB.

The correct option is C 60∘In△OCDOC=OB=CD∴∠COD=60(Anglesofequilateraltriangle)AS∠CAD=12∠COD=12×60=30Now∠ADP=∠ADB=90(Angleinsemicircle)In△ADP∠ADP+∠DPA+∠DAP=180⇒90+∠DPA+30=180⇒∠DPA=180−120=60=∠APBHenceoption(c)isrightanswer

In the given figure, $A B$ is the diameter of a circle $C (O, r)$ . chord $C D$ is equal to radius $O C$ . If $A C a n d B D$ intersect at $P$ , find $∠ A P B$ .