Question

if sin(A+2B)=√3/2 and cos (A+4B)=0, 0°<A, B<90°,A>B, find A and B​

Answer 1

Answer:

A = 30°

B = 15°

Step-by-step explanation:

sin ( A + 2B ) = √3 / 2

=> sin ( A + 2B) = sin 60°

=> A + 2B = 60° ---- Eq( 1 )

cos ( A + 4B ) = 0

=> cos ( A + 4B ) = cos 90°

=> A + 4B = 90° --- Eq( 2 )

Subtrating Eq( 1 ) From Eq( 2 )

=> A + 4B - ( A + 2B ) = 90° - 60°

=> A + 4B - A - 2B = 30°

=> 2B = 30°

=> B = 30° / 2 = 15°

Substituting B = 15° in Eq( 1 )

=> A + 2( 15 ) = 60°

=> A + 30° = 60°

=> A = 60° - 30°

=> A = 30°

Answer 2

Answer:

Hope it works....................™✌️✌️