Question

If sin A = 3/4, calculate cos A and tan A..​

Answer 1

sin A = 3/4 = Perpendicular/Hypotenuse

Let Base be B, Perpendicular be P and hypotenuse be H

Using Pythagoras Theorem,

H^2 = B^2 + P^2

4^2 = B^2 + 3^2

16 = B^2 + 9

B^2 = 16-9

B = √ 7

cos A = B/H = √7/4

tan A = P/B = 3/√7

Answer 2

Given that,

• $\rm { sin \: A = \dfrac{3}{4} }$

We have to find,

• $\rm { cos \: A \: and \: tan \: A }$

Solution,

Here, we know that

$\rm { sin \: A =\dfrac{ 3}{4} = \dfrac{ Perpendicular}{Hypotenuse} }$

Hence,

• $\rm { Perpendicular = 3}$
• $\rm { Hypotenuse = 4}$

_____________

Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2} + {P}^{2} }$

$\rm { \leadsto {B}^{2} = {H}^{2} - {P}^{2} }$

$\rm { \leadsto {B}^{2} = {4}^{2} - {3}^{2} }$

$\rm { \leadsto {B}^{2} = 16 - 9 = 7}$

$\rm\blue { \leadsto B = \sqrt{7} }$

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

$\leadsto \rm\red{ cos \: A = \dfrac{ Base}{ Hypotenuse} = \dfrac{ \sqrt{7} }{4} }$

$\:$

$\leadsto \rm\red{ tan \: A = \dfrac{ Perpendicular}{ Base} = \dfrac{ 3 }{\sqrt{7}} }$