Question

in the given figure,O is the centre of the circle. determine angle APC,if DC and DC are tangents and angle ADC=50 degree.(5marks).

Answer 1

$\huge{\underline{\underline{\blue{\mathfrak{Answer :}}}}}$

★ Given :

$\sf{ \angle{ADC} = {50}^{ \circ} } \\ \sf{ad \: and \: dc \: are \: tangents}$

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★ To Find :

$\sf{To \: find \: \angle{APC}}$

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★ Solution :

Join AO and CO , To form a quadrilateral AOCD.

As, Tangent of a circle are perpendicular to the radius at point of contact.

So,

$\sf{ \angle{DAO = \angle{DCO} = {90}^{ \circ} }}$

Now, in quadrilateral AOCD

$\sf{ \angle{D} + \angle{A} + \angle{O} +\angle{C} = {360}^{ \circ} } \\ \\ \sf{ {50}^{ \circ} + {90}^{ \circ} + \angle{O} + {90}^{ \circ} = {360}^{ \circ} } \\ \\ \sf{ {230}^{ \circ} + \angle{O} = {360}^{ \circ} } \\ \\ \sf{\angle{O} = {360}^{ \circ} - {230}^{ \circ} } \\ \\ \sf{\angle{O} = {130}^{ \circ} }$

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Now,

By using "Central Angle Theorem"

$\sf{\angle{APC} = \frac{ {360}^{ \circ} - \angle{O} }{2} } \\ \\ \sf{\angle{APC} = \frac{ {360}^{ \circ} - {130}^{ \circ} }{2} } \\ \\ \sf{\angle{APC} = \frac{ { \cancel{230}}^{ \circ} }{ \cancel2} } \\ \\ { \LARGE{ \boxed{ \orange{\sf{\angle{APC} = {115}^{ \circ}}}}}}$

Answer 2

Answer:

We know, the angle between tangent and radius at the point of contact is a right angle.

               ∴∠OAD=∠OCD=90

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               Applying angle sum property in quadrilateral OADC, we get,

               ⇒∠OAD+∠ADC+∠OCD+∠AOC=360

∘

               ⇒90

∘

+50

∘

+90

∘

+∠AOC=360

∘

               ⇒230

∘

+∠AOC=360

∘

               ⇒∠AOC=130

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Step 3: Find the required value using suitable property.

               Now, Reflex ∠AOC=360

∘

−130

∘

                                                     =230

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               We know, angle subtended at remaining part of a circle is half of the angle subtended at centre.

               ∴∠APC=

2

1

​

 ∠AOC

                                 =

2

1

​

×230

∘

                                 =115

∘

Hence, the required measure is 115