9. Given secA
13
then find the value of sin A.
12
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Step-by-step explanation:
Let △ABC be right angled triangle (right angled at B)
secQ=
AB
AC
=
12
13
Let AC=13x and AB=12x
AC
2
=AB
2
+BC
2
BC
2
=AC
2
−AB
2
x
2
=169−144
x
2
=25
x=
25
=5
sinQ=
AC
BC
=
13x
5x
=
13
5
cosQ=
AC
AB
=
13
12
tanQ=
AB
BC
=
12x
5x
=
12
5
cotQ=
tanQ
1
=
12
5
1
=
5
12
cosec Q=
BC
AC
=
5x
13x
=
5
13
soluti
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1
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