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Ex 5.5
6. find the length of the chord cut off by an angle of π/6 subtended at the centre of a circle with radius 5cm

Sagot :

crisforp
;
2 × 5 × sin( π / 12 ) = 10 × sin ( π / 12 ) ;

π/12 = ( 180° ) / 12 = 15°. 

But,  sin ( π/12 ) 

= sin 15° 

= sin ( 45° - 30°) 

= sin 45°· cos 30° - cos 45°· sin 30° 

= (1/√2)·(√3 /2 ) - (1/√2)·(1/2) 

= ( √3 - 1 ) / (2√2) = [tex] \frac{( \sqrt{6}- \sqrt{2} )}{4} [/tex] ;

Finally, ( 2.5 ) × ( 2.4 - 1.4 ) = 2.5 cm ;
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