Question

If sin A = 3/4 then find cos A ​
​

Answer 1

Answer:

Given, sinA=

4

3

⇒

DC

BC

=

4

3

⇒BC=3k and AC=4k

where k is the constant of proportionality.

By Pythagoras theorem, we have

AB

2

=AC

2

−BC

2

=(4k)

2

−(3k)

2

=7k

2

⇒AB=

7

k

So, cosA=

AC

AB

=

4k

7

k

=

4

7

And tanA=

AB

BC

=

7

k

3k

=

7

3

Step-by-step explanation:

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Answer 2

Given :-

• sin A = ¾

To Find :-

• Value of cos A

Formula to be used :-

• sin² x + cos² x =1

Solution :-

Given that,

sin A = ¾

We know,

$\rm{\sin^2x+\cos^2x=1}$

$\implies\rm{\cos\:A=\sqrt{1-\sin^2A}}$

$\implies\rm{\cos\:A=\sqrt{1-(\dfrac{3}{4})^2}}$

$\implies\rm{\cos\:A=\sqrt{1-\dfrac{9}{16}}}$

$\implies\rm{\cos\:A=\sqrt{\dfrac{16-9}{16}}}$

$\implies\rm{\cos\:A=\dfrac{\sqrt{7}}{4}}$

Hence, value of cos A is = √7/4

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Some Formulas :-

• sin² x + cos² x =1

• 1 + tan²x = sec²x

• cos²x - sin²x = cos 2x

• 1 + cot²x = cosec²x