Question

given that: Sin A= a/b, find cos A.​

Answer 1

$\large\green{\boxed{ANSWER}}$

For figure refer to attachment:

GIVEN:

$\large\red{\boxed{Sin A=\dfrac {a}{b}}}$

TO FIND:

$cosA$

SOLUTION:

We know

sinA =$\dfrac{perpendicular}{hypotenuse}$

Let AB be s.

In right ∆ ABC,

$\large\orange{\boxed{AC^{2}= BC^{2}+AB^{2}}}$

$\large\blue{(By \:Pythagoras\:Theorem)}$

=>$AB^{2}=AC^{2}-BC^{2}$

=>$s^{2}=b^{2}-a^{2}$

=>$s =\sqrt{b^{2}-a^{2}}$

=>Therefore $s= \sqrt{b^{2}-a^{2}}$

We know,

$cosA =\dfrac {base}{hypotenuse}$

= >$cosA =\dfrac{\sqrt{b^{2}-a^{2}}}{b}$

Hence $\large\purple{\boxed{cosA =\dfrac{\sqrt{b^{2}-a^{2}}}{b}}}$