Question

$\huge \fbox \red{❥ Question}$


If Sin A = 3/4, Calculate cos A and tan A.


✳️gud luck :-P​

Answer 1

cosA = √7/4 & tanA = 3/√7

If sinA = 3/4 , then it the opposite side of angle A in the right triangle will be 3 & Hypotenuse of the triangle will be 4 . So we should find the adjacent side of angle A .

So , using Pythagoras theorem

◕ Hypotenuse² = Base² + Height²

→ 4² = Base² + 3²

→ 16 - 9 = Base²

→ Base = √7 = Adjacent

Now ,

✧ cosA = Adjacent/Hypotenuse = √7/4

✧ tanA = opposite/Adjacent = 3/√7

Answer 2

${\huge{\rm{\underline{\underline{Question:-}}}}}$

→ If Sin A = 3/4, Calculate cos A and tan A

${\huge{\rm{\underline{\underline{Given:-}}}}}$

$\sf Sin\; A = \dfrac{3}{4}$

${\huge{\rm{\underline{\underline{To\;Find:-}}}}}$

→  cos A and tan A

${\huge{\rm{\underline{\underline{Solution:-}}}}}$

→ As , we're given that

$\sf sin\;A = \dfrac{3}{4} \\\\\\\ \rightarrow \dfrac{BC}{DC} = \dfrac{3}{4} \\\\\\\rightarrow BC = 3k \; and\; AC = 4k\\\\\\\longrightarrow where\;k\;is\;the\;constant\;of\;proportionality.$

→ By Pythagoras theorem :

${\boxed{ \underline{AB^{2}= AC^{2} - BC^{2}}}$

→ let's solve !!

$\sf \longrightarrow (4k)^{2} - (3k^{2}) = 7k^{2} \\\\\longrightarrow AB = \sqrt{7k} \\\\\\\longrightarrow cos A = \dfrac{AB}{AC} = \dfrac{\sqrt{7k} }{4k} = \dfrac{\sqrt{7} }{4} \\\\\\\longrightarrow tan A = \dfrac{BC}{AB} = \dfrac{3k}{\sqrt{7k}} = \dfrac{3}{\sqrt{7}}$

__________

All Done :D !!