Sol.:iv) If cos A = 15/17 then find sin A.Sol.:

1.	Sol.:iv) If cos A = 15/17 then find sin A.Sol.:
Answer» ${\boxed{\underline{\tt{\orange{Required \: \: answer:-}}}}}$ ★GIVEN:- $\displaystyle \sf{Cos A = \frac{15}{17} }$ ★TO FIND:- $\displaystyle \sf{<klux>SIN</klux> A }$ ★FORMULA USED:- PHYTHAGORAS theorem:- $\displaystyle \sf{HYPOTENUSE^2 = BASE^2 +PERPENDICULAR^2}$ And, $\displaystyle \sf{Cos A = \frac{Base }{Hypotenuse} }$ $\displaystyle \sf{Sin A = \frac{Perpendicular}{Hypotenuse} }$ ★SOLUTION:- As we know that, Cos A = B/H It MEANS that Base = 15 and Hypotenuse = 17 We will find Perpendicular by the use of phythagoras theorem:- $\displaystyle \sf : \longrightarrow{ {(17)}^{2} = {(15)}^{2} + {P}^{2} } \\ \\\displaystyle \sf : \longrightarrow{ 289= 225+ {P}^{2} } \\ \\ \: \displaystyle \sf : \longrightarrow{ {P}^{2} = 64} \\ \\ \displaystyle \sf : \longrightarrow{ {P}^{2} = {<klux>8</klux>}^{2} } \\ \\ \sf{on \: comparing \:as \: the \: powers \: are \: equal} \\ \\ \displaystyle \sf : \longrightarrow{ \pink{P= {8}}}$ Now the Perpendicular is 8. So, $\displaystyle \sf{Sin A = \frac{Perpendicular}{Hypotenuse} }$ ${ \boxed{ \underline{ \purple{ \sf{ \bf{{Sin \: A = \frac{8}{17} }}}}}}}$ HENCE, the value of Sin A = 8/17.

Answer»

${\boxed{\underline{\tt{\orange{Required \: \: answer:-}}}}}$

★GIVEN:-

$\displaystyle \sf{Cos A = \frac{15}{17} }$

★TO FIND:-

$\displaystyle \sf{<klux>SIN</klux> A }$

★FORMULA USED:-

PHYTHAGORAS theorem:-

$\displaystyle \sf{HYPOTENUSE^2 = BASE^2 +PERPENDICULAR^2}$

And,

$\displaystyle \sf{Cos A = \frac{Base }{Hypotenuse} }$

$\displaystyle \sf{Sin A = \frac{Perpendicular}{Hypotenuse} }$

★SOLUTION:-

As we know that,

Cos A = B/H

It MEANS that Base = 15 and Hypotenuse = 17

We will find Perpendicular by the use of phythagoras theorem:-

$\displaystyle \sf : \longrightarrow{ {(17)}^{2} = {(15)}^{2} + {P}^{2} } \\ \\\displaystyle \sf : \longrightarrow{ 289= 225+ {P}^{2} } \\ \\ \: \displaystyle \sf : \longrightarrow{ {P}^{2} = 64} \\ \\ \displaystyle \sf : \longrightarrow{ {P}^{2} = {<klux>8</klux>}^{2} } \\ \\ \sf{on \: comparing \:as \: the \: powers \: are \: equal} \\ \\ \displaystyle \sf : \longrightarrow{ \pink{P= {8}}}$

Now the Perpendicular is 8.

So,

$\displaystyle \sf{Sin A = \frac{Perpendicular}{Hypotenuse} }$

${ \boxed{ \underline{ \purple{ \sf{ \bf{{Sin \: A = \frac{8}{17} }}}}}}}$

HENCE, the value of Sin A = 8/17.