Question

Solve it!! sin(A+2B)=√3/2,cos(2A-B)=1/√2​

Answer 1

Step-by-step explanation:

Given, sin(A+2B)=

2

3

cos(A+4B)=0

A>B,

We know that, sin60

∘

=

2

3

and cos90

∘

=0

Consider,

sin(A+2B)=

2

3

and sin60

∘

=

2

3

⟹(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=

2

30

∘

=15

∘

From (ii)

(A+4B)=90

∘

Also, B=15

∘

A=90

∘

−60

∘

A=30

∘

∴A=30

∘

,B=15

∘