Math, asked by rovaljotk, 4 months ago

9. Given secA
13
then find the value of sin A.
12

Answers

Answered by gudduchoudhary1983

0

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

Answered by neelamthakur2100

1

Answer:

hope it's helpful to you

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