Question

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

Answer 1

SOLUTION
Let ΔABC be a right-angled triangle, right-angled at B. We know that sin A = BC/AC = 3/4 Let BC be 3k and AC will be 4k where k is a positive real number. By Pythagoras theorem we get, AC2 = AB2 + BC2 (4k)2 = AB2 + (3k)2 16k2 - 9k2 = AB2 AB2 = 7k2 AB = √7 k cos A = AB/AC = √7 k/4k = √7/4 tan A = BC/AB = 3k/√7 k = 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,

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

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

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

$\:$

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