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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 26, 2021
 #1
avatar+2135 
0

let's draw a triangle, with the legs, 1 and 2. 

So th hypotenues is sqrt(5)

Sine = 2/sqrt(5)

Cos = 1/sqrt(5)

 

=^._.^=

 Apr 26, 2021
 #2
avatar+1452 
+1

sinA = 2(cosA)

 

sin(A) / cos(A) = tan(A)

 

tan(A) = 2        ∠A = 63.43494882º

 

cos(A) = 0.447213595

 Apr 27, 2021

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