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

 Apr 30, 2022
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\($\sin A = 2\cos A$\)

 

Note  that  cos A     ≠  0   ( A would have to be 90°....but we know  that B  = 90°)

 

So  we  have  that

 

sin A   / cos A   = 2

 

tan A  = 2

 

Which implies   that

 

y = 2    and x  =1

 

r  =  sqrt [ 1^2 + 2^2 ]   =sqrt 5

 

So

 

cos A   = x  / r   =   1  /  sqrt 5  =     sqrt (5)  /  5

 

 

cool cool cool

 Apr 30, 2022

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