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Standard X

Mathematics

Extension of Inscribed Angle Theorem

In the follow...

Question

In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle. If $m (a r c P C R) = 26^{o}, m (a r c Q D S) = 48^{o}$ . Then find :

Measure of $∠ P Q R$ (in degrees)

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Solution

Here,
$m (a r c P C R) = 26^{o}, m (a r c Q D S) = 48^{o}$
$∠ P Q R$ intercepts arc PCR.
$∴ ∠ P Q R = \frac{1}{2} m (a r c P C R)$ (By Inscribed Theorem)
$∴ m ∠ P Q R = \frac{1}{2} \times 26^{o} = 13^{o}$

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Q. Secants containing chords RS and PQ of a circle intersects each other at point A in the exterior of a circle as shown in the given figure.
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Q. In the given figure, PQ and RS are two chords of a circle intersecting each other at a point T outside the circle. If PT = 3 cm, PQ = 5 cm, RS = 2 cm, then the length of RT is

In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD $⊥$ PQ and OE $⊥$ RS. If $∠ D O E = 130^{\circ},$ then $∠ P D E$ (in degrees) = ___.

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In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle. If m(arcPCR)=26o,m(arcQDS)=48o. Then find :Measure of ∠PQR (in degrees)

Here, m(arcPCR)=26o,m(arcQDS)=48o∠PQR intercepts arc PCR.∴∠PQR=12m(arcPCR) (By Inscribed Theorem)∴m∠PQR=12×26o=13o

In the following figure, secants containing chords RS and PQ of a circle intersects each other in point A in the exterior of a circle. If $m (a r c P C R) = 26^{o}, m (a r c Q D S) = 48^{o}$ . Then find :

Measure of $∠ P Q R$ (in degrees)

Here,
$m (a r c P C R) = 26^{o}, m (a r c Q D S) = 48^{o}$
$∠ P Q R$ intercepts arc PCR.
$∴ ∠ P Q R = \frac{1}{2} m (a r c P C R)$ (By Inscribed Theorem)
$∴ m ∠ P Q R = \frac{1}{2} \times 26^{o} = 13^{o}$