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In right triangle ABC with \(\angle B = 90^\circ\), we have \(\sin A = 2\cos A\). What is \(\cos A\)?

 Jun 26, 2019
 #1
avatar+19913 
0

Sin a = 2 cos a

sin a / cos a = 2

arctan 2 = 63.44 degrees

 

 

Ooops sorry.... I gave ypu the angle, not the cos          

cos 63.44 = .447      

 

Chris' answer below is the EXACT value.....     ~ EP

 Jun 26, 2019
edited by ElectricPavlov  Jun 26, 2019
edited by ElectricPavlov  Jun 26, 2019
 #2
avatar+106533 
+2

sin A   =   2cosA

 

sin^2A  + cos^2A  = 1

 

[ 2cosA]^2 + cos^2A  = 1

 

4cos^2 A  + cos^2A  = 1

 

5cos^2A  =  1

 

cosA^2 A  =   1 /5

 

cosA =  ±√[1/5]

 

But....since this is a right triangle, the cosine could only be negative if angle A was obtuse which is impossble

 

So

 

cos A  = √[1/5]

 

 

 

cool cool cool

 Jun 26, 2019

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