Question

$\large\underline{\sf{question-}}$
What is the value of
☞If Sin A = 3/4, Calculate cos A and tan A.
​

Answer 1

Answer:

$\sin \: a = \frac{3}{4}$

$\frac{side \: opposite \: to \: \angle \: a}{hypotenuse} = \frac{3}{4}$

$\frac{bc}{ac} = \frac{3}{4}$

Let BC=3x

AC=4x

Let's find AB by pythagoras theorem.

${(hypotenuse)}^{2} = {(height)}^{2} + {(base)}^{2}$

${ac}^{2} = {ab}^{2} + {bc}^{2} \\ \\ {(4x)}^{2} = {ab}^{2} + {(3x)}^{2} \\ \\ {16x}^{2} = {ab}^{2} + {9x}^{2} \\ \\ {16x}^{2} - {9x}^{2} = {ab}^{2} \\ \\ {7x}^{2} = {ab}^{2} \\ \\ {ab}^{2} = {7x}^{2} \\ \\ ab = \sqrt{ {7x}^{2} } \\ \\ ab = \sqrt{7} x$

Now we need to find Cos A and tan A

So first let's find out cos A

$\cos \: a = \frac{side \: adjacent \: to \angle \: a}{hypotenuse} \\ \\ = \frac{ab}{ac} \\ \\ = \frac{ \sqrt{7} x}{4x} \\ \\ = \frac{ \sqrt{7} }{4}$

Now tan A

$\tan \: a = \frac{side \: opposite \: to \: \angle \: a}{side \: adjacent \: to \: \angle \: a}$

$\tan \: a = \frac{bc}{ab} \\ \\ tan \: a = \frac{3x}{ \sqrt{7}x } \\ \\ tan \: a = \frac{3}{ \sqrt{7} }$

Hope it helps you from my side

Answer 2

Given :

• ${\sf{Sin~A~=~\dfrac{3}{4}}}$

To Find :

• ${\sf{Find~ \cos~A~and~ \tan~A~\bf{?}}}$

______________

Solution : Sin A & tan A.

$~$

• $\boxed{\sf{\pink{\pmb{Sin~A~=~\dfrac{3}{4}~=~\dfrac{Perpendicular}{Hypotenuse}}}}}$★

$~$

Therefore,

• Perpendicular = 3
• Hypotenuse = 4

$~$

${\bf{\underline{\pmb{Applying~ Pythagoras~ Property~:}}}}$

$~$

• $\boxed{\sf{\purple{\pmb{H^2~=~B^2~+~P^2}}}}$★

$~$

${\sf:\implies{H^2~=~B^2~-~P^2}}$

${\sf:\implies{B^2~=~4^2~-~3^2}}$

${\sf:\implies{B^2~=~16~-~9}}$

${\sf:\implies{B^2~=~7}}$

• ${\underline{\boxed{\pmb{\sf{\pink{B~=~\sqrt{7}}}}}}} \; {\red{\bigstar}}$

$~$

______________

Hence,

• ${\rm\rightarrow{Hypotenuse~=~4}}$
• ${\rm\rightarrow{Base~=~\sqrt{7}}}$
• ${\rm\rightarrow{Perpendicular~=~3}}$

$~$

______________

• ${\sf{\cos~A~=~\dfrac{Base}{Hypotenuse}~=~\bf{\pmb{\dfrac{\sqrt{7}}{4}}}}}$

• ${\sf{\tan~A~=~\dfrac{Perpendicular}{Base}~=~\bf{\pmb{\dfrac{3}{\sqrt{7}}}}}}$