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In Fig.23, O A\ a n d\ O B are opposite ...

In Fig.23, `O A\ a n d\ O B` are opposite rays: (i) If `( i i ) (iii) x=( i v ) (v) 25^(( v i )0( v i i ))( v i i i ) , (ix)` (x) what is the value of `( x i ) (xii) y ? (xiii)` (xiv) (xv) If `( x v i ) (xvii) y=( x v i i i ) (xix) 35^(( x x )0( x x i ))( x x i i ) , (xxiii)` (xxiv) what is the value of `( x x v ) (xxvi) x ? (xxvii)` (xxviii)

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Convert the following polar coordinates to its equivalent Cartesian coordinates. (i) ( i i ) (iii)2,pi)( i v ) (v) (ii) ( v i ) (vii)(( v i i i ) (ix)sqrt(( x )2( x i ))( x i i ) , (xiii)pi/( x i v )6( x v ) (xvi) (xvii))( x v i i i ) (xix) (iii) ( x x ) (xxi)(( x x i i ) (xxiii)-3,-( x x i v )pi/( x x v )6( x x v i ) (xxvii) (xxviii))( x x i x ) (xxx)

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In Fig.78, we have (i) ( i i ) (iii)/_M L Y=2/_L M Q , (iv) (v) find ( v i ) (vii)/_L M Q (viii) (ix) (x) ( x i ) (xii)/_X L M=( x i i i ) (xiv)(( x v ) (xvi)2x-10 (xvii))^(( x v i i i )0( x i x ))( x x ) (xxi) (xxii) and ( x x i i i ) (xxiv)/_L M Q=x+( x x v ) (xxvi) 30^(( x x v i i )0( x x v i i i ))( x x i x ) , (xxx) (xxxi) find ( x x x i i ) (xxxiii) xdot( x x x i v ) (xxxv) (xxxvi) ( x x x v i i ) (xxxviii)/_X L M=/_P M L ,\ ( x x x i x ) (xl) find ( x l i ) (xlii)/_A L Y (xliii) (xliv) (xlv) ( x l v i ) (xlvii)/_A L Y=( x l v i i i ) (xlix)(( l ) (li)2x-15 (lii))^(( l i i i )0( l i v ))( l v ) ,\ /_L M Q=( l v i ) (lvii)(( l v i i i ) (lix) x+40 (lx))^(( l x i )0( l x i i ))( l x i i i ) , (lxiv) (lxv) find ( l x v i ) (lxvii) xdot( l x v i i i ) (lxix)

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