. Find acute angles A and B if sin (A + 2B) =√3/2
and cos (A + 4B) = 0, A > B, 0º < A + 2B ≤ 90°,
0° <A + 4B ≤ 90°.
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:
Question:-
- Find acute angles A and B if sin (A + 2B) =√3/2 and cos (A + 4B) = 0, A > B, 0º < A + 2B ≤ 90°, 0° < A + 4B ≤ 90°.
Answer :-
Concept used :-
- Trigonometric Ratios of Standard angles
Given :-
- sin (A + 2B) =√3/2 and cos (A + 4B) = 0,
- A > B, 0º < A + 2B ≤ 90°, 0° <A + 4B ≤ 90°.
To Find :-
- Values of acute angles A and B
Solution:-
☆ As sin (A + 2B) =√3/2
⟹ sin(A + 2B) = sin 60°
On comparing, we get
⟹ A + 2B = 60° ..............[1]
☆Also, cos(A + 4B) = 0
⟹ cos(A + 4B) = cos90°
On comparing, we get
⟹ A + 4B = 90° ............[2]
☆Now, Subtracting [1] from [2], we get
A + 4B - A - 2B = 90° - 60°
⟹ 2B = 30°
⟹ B = 15°
☆Put the value of B in equation [1] we get
A + 2 × 15° = 60°
⟹ A = 60° - 30°
⟹ A = 30°
Hence, acute angle A = 30° and B = 15°.