In ∆ABC, right angled at B, if AB = 12 cm and BC = 5 cm, find
i) sin A and tan A ii) sin A and cot A.
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1
AC
thus
=
12
2
+5
2
=13
sinA=
AC
BC
=
13
5
tanA=
AB
BC
=
12
5
sinC=
AC
AB
=
13
12
cotC=
AB
BC
=
12
5
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