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# the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians.

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the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians.

a) find the length of the arc PQ, giving the answer as a multiple of π

b) the tangents to the cirlce at the point P and Q meet at the point T. and the angles TPO AND TQO are both right angles

i) Angle PTQ = ∝ radians. find ∝ in terms of pi

ii)Find the area of the shaded region bounded by the arc PQ and the tangents TP and TQ

Aug 16, 2019

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the diagram shows a sector OPQ of a circle with centre O. the radius of the circle is 18m and the angle POQ is 2π / 3 radians.

a) find the length of the arc PQ, giving the answer as a multiple of π

b) the tangents to the cirlce at the point P and Q meet at the point T. and the angles TPO AND TQO are both right angles

i) Angle PTQ = ∝ radians. find ∝ in terms of pi

ii)Find the area of the shaded region bounded by the arc PQ and the tangents TP and TQ

a)   arc length PQ  =  18 *  (2 pi)/3 =     36 pi / 3   =   12 pi

b) i)   Angle  PTQ   will be  supplemental to angle POQ   =   pi  - (2pi)/3   =  pi/3   = 60°

ii )    The area of the circular sector  formed by  arc PQ  =  (1/2)r^2 (2pi/3)  =  (1/2)(18)^2 (2pi)/3 = 108 pi  m^2     (1)

[PTQO]   will from a  kite   ...OQ  = 18     and  triangle TPO will form a 30-60-90 right  triangle with QO = 18 opposite the 30° angle  (angle QTO)  and QT  opposite the 60° angle  (angle QOT)....so  QT  = 18√3

So  the area of this kite =  product of the legs  of triangle OQT =  QO * QT  = 18 * 18√3  = 324√3    (2)

So...the area bounded by arc PQ  and the  tangents TP and TQ   = (2)  - (1)  =  324√3 - 108 pi ≈  221.9 m^2   Aug 16, 2019