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Exercise (11.3) - No(1) Sample Solutions


(b) Solution (1)

sin135=sin(18045)=sin45=22

cos135 =cos(18045)=cos45=22

tan135 =1cot135 =1sec135 =2cosec135 =2

(b) Solution (2)

   principal angle = 135basic acute angle = 45

 sin45=22and cos45=22

   sin135=sin45and cos135=cos45 numerically.   But 135 lies in the second quadrant.

sin135=sin45=22

   cos135=cos45=22

   tan135 =1cot135 =1sec135 =2cosec135 =2

(d) Solution (1)

   sin210=sin(180+30)=sin30=12

   cos210=cos(180+30)=cos30=32

   tan210=33

   cot210=3

   sec210=233

   cosec210=2

(d) Solution (2)

   principal angle = 210

basic acute angle = 30

 sin30=12and cos30=32

   sin210=sin30and cos210=cos30 numerically.

   But 210 lies in the third quadrant.

sin210=sin30=12

   cos210=cos30=32

   tan210 =33cot210 =3sec210 =233cosec210 =2

(m) Solution (1)

   sin(120)=sin120

                   =sin(18060)

                   =sin60

                   =32

    cos(120)=cos120

                     =cos(18060)

                     =cos60

                     =12

   tan(120)=3

   cot(120)=33

   sec(120)=2

   cosec(120)=233

(m) Solution (2)

   principal angle = 120

basic acute angle = 60

 sin60=32and cos60=12

   sin(120)=sin60and cos(120)x=cos60 numerically.

   But (120) lies in the third quadrant.

sin(120)=sin60=32

   cos(120)=cos60=12

   tan(120)=3

   cot(120)=33

   sec(120)=2

   cosec(120)=233

(q) Solution (1)

   sin480=sin(360+60)
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               =sin60

               =32

    cos480=cos(360+60)

                 =cos60

                 =12

   tan480=3

   cot480=33

   sec480=2

   cosec480=233

(q) Solution (2)

   principal angle = 480

basic acute angle = 60

 sin60=32and cos60=12

   sin480=sin60and cos480=cos60 numerically.

   But 480 lies in the first quadrant.

sin480=sin60=32

   cos480=cos30=12

   tan480=3

   cot480=33

   sec480=2

   cosec480=233


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