Q. 184.3( 29 Votes )

If the angle betw

Answer :

2r



Given - BPA = 60⁰


To find - length of OP


Properties -


1. Lengths of the two tangents drawn from an external point to a circle are equal.


2. In 30⁰-60⁰-90⁰ triangles, the side opposite to 30⁰ is half of the hypotenuse and the side opposite to 60⁰ is times the hypotenuse.


3. The radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.


Answer -


As we know that lengths of tangents drawn from the same external point to the circle are same.


PA = PB ………(1)


The angle between the two tangents is 60⁰.


BPA = 60⁰ ………(2)


Therefore, for ∆POB & ∆POA,


PB = PA ………from (1)


OB = OA = r ………..radii of same circle


And OP is the common side.


………by SSS test of similarity


BPO = APO ……….corresponding angles of similar triangles ……….(3)


But BPA = BPO + APO


60⁰ = BPO + BPO ……….from (2) & (3)


60⁰ = 2 . BPO


BPO = 30⁰ = APO


As tangents are always perpendicular to the radius


PBO = PAO = 90⁰


Therefore, ∆POB & ∆POA are 30⁰-60⁰-90⁰.


Therefore, by the property of 30⁰-60⁰-90⁰ triangle, the side opposite to 30⁰ is half of hypotenuse.


In ∆POB, BPO = 30⁰ and side opposite to BPO is OB, therefore,




OP = 2r


Hence, the length of OP is 2r units.


OR


6 cm


In the figure below, EF is the chord for outer circle, and it is tangent for inner circle at point D.



Given – radius of inner circle = AD = 4 cm


Radius of outer circle = AE = AF = 5 cm


To Find – length of chord EF


Properties –


1. The radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.


Answer –


As radius from the centre of the circle to the point of tangency is perpendicular to the tangent line.


ADE = ADF = 90⁰ ………(1)


AE = AF = 5 cm ………(2) (given)


AD = 4 cm ………(3) (given)


In right angled triangle ADE,


By Pythagoras Theorem,


AE2 = AD2 + ED2


52 = 42 + ED2


25 = 16 + ED2


ED2 = 9


ED = 3 ………(4)


Similarly, in right angled triangle ADF,


AF2 = AD2 + FD2


52 = 42 + FD2


25 = 16 + FD2


FD2 = 9


FD = 3 ………(5)


Now, EF = ED + FD


EF = 3 + 3 = 6 cm


Therefore, length of chord is 6 cm.


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