In the given figure, $A B$ and $C D$ are equal chords of a circle, with centre $O$ . If $P$ is the mid-point of chord $A B, Q$ is the mid-point of chord $C D$ and $∠ P O Q = 150^{o}$ find $∠ A P Q$ .

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Solution

$\begin{matrix} E q u a l c h o r d s a r e s i t u a t e d a t e q u a l d i s tan c e O P = O Q I n a t r i a n g l e w h e n t w o c h o r d s a r e e q u a l s i t s a n g l e a r e e q u a l ∠ P = ∠ Q = x s u m o f a n g l e o f Δ O P Q 150 + x + x = 180^{0} 2 x = 180^{0} - 150^{0} = 30^{0} x = \frac{30}{2} = 15^{0} ∠ A P O = x + ∠ A P Q = 90^{0} ∠ A P Q = 90^{0} - 15^{0} = 75^{0} \end{matrix}$

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In the given figure, AB and CD are equal chords of a circle, with centre O. If P is the mid-point of chord AB,Q is the mid-point of chord CD and ∠POQ=150o find ∠APQ.

EqualchordsaresituatedatequaldistanceOP=OQInatrianglewhentwochordsareequalsitsangleareequal∠P=∠Q=xsumofangleofΔOPQ150+x+x=18002x=1800−1500=300x=302=150∠APO=x+∠APQ=900∠APQ=900−150=750

In the given figure, $A B$ and $C D$ are equal chords of a circle, with centre $O$ . If $P$ is the mid-point of chord $A B, Q$ is the mid-point of chord $C D$ and $∠ P O Q = 150^{o}$ find $∠ A P Q$ .