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Question

Find acute angles A and B, if sin(A+2B)=32 and cos(A+4B)=0, and A>B.

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Solution

Given, sin(A+2B)=32
cos(A+4B)=0
A>B,

We know that, sin60=32 and cos90=0

Consider,
sin(A+2B)=32 and sin60=32

(A+2B)=60 ---------------(i)

Consider,
cos(A+4B)=0 and cos90=0

(A+4B)=90 ---------------(ii)

Solve (i) and (ii) :

(A+2B)=60
(A+4B)=90
Subtracting (i) from (ii),

2B=30

B=302=15

From (ii)
(A+4B)=90

Also, B=15
A=9060
A=30

A=30,B=15

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