The given figure shown a circle with centre O and $∠$ ABP = $42^{o}$ .

Calculate the measure of :

(i) $∠$ PQB

(ii) $∠$ QPB + $∠$ PBQ

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Solution

Join A and P as shown in the diagram.

(i) $\angle APB = 90^o$ (angle in semi circle)

$\therefore \angle BAP = 90^o-\angle ABP = 90^o-42^o = 48^o$

Now, $\angle PQB = \angle BAP = 48^o$ (angles in the same segment)

(ii) in $\triangle BPQ$

$\angle QPB + \angle PBQ = 180^o-\angle PQB = 180^o-48^o = 132^o$

$\\$

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The given figure shown a circle with centre O. Also, PQ = QR = RS and $∠$ PTS = $75^{o}$ . Calculate :

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In the given figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is

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