13
Find A and B, if sin (A + B) =
cos (A - B), (where A and B are acute angles).
Answers
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
∘
Answer:
Given, sin(A+2B)=23
cos(A+4B)=0
A>B,
We know that, sin60∘=23 and cos90∘=0
Consider,
sin(A+2B)=23 and sin60∘=23
⟹(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=230∘=15∘
From (ii)
(A+
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