Question
Find acute angles A and B, if $\sin(\text{A + 2B})=\frac{\sqrt{3}}{2}$ and $\cos(\text{A + 4B})=0,\text{A}>\text{B}.$
✓
Answer
We have,
$\sin(\text{A}+2\text {B})=\frac{\sqrt {3}}{2}$ and $\cos(\text {A}+4\text{B})=0$
Now, $\sin(\text{A}+2\text {B})=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin (\text{A}+2\text{B})=\sin60^\circ$
$\Rightarrow\text {A}+2\text{B}=60^ \circ\dots(1)$
And, $\cos(\text{A}+4\text {B})=0$
$\Rightarrow\cos (\text{A}+4\text{B})= \cos90^\circ$
$\Rightarrow\text {A}+4\text{B}=90^ \circ\dots(2)$
Substract (1) from (2), we get
$\Rightarrow2\text{B} =30^\circ$
$\Rightarrow\text{B}= \frac{30^\circ} {2}=15^\circ$
Put B = 15° in (1) we get -
$\Rightarrow\text {A}+2\times15=60^ \circ$
$\Rightarrow\text {A}+30^\circ=60^\circ $
$\Rightarrow\text{A} =60^\circ-30^\circ $
$\Rightarrow\text{A} =30^\circ$
Thus, A = 30° and B = 15°
$\sin(\text{A}+2\text {B})=\frac{\sqrt {3}}{2}$ and $\cos(\text {A}+4\text{B})=0$
Now, $\sin(\text{A}+2\text {B})=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin (\text{A}+2\text{B})=\sin60^\circ$
$\Rightarrow\text {A}+2\text{B}=60^ \circ\dots(1)$
And, $\cos(\text{A}+4\text {B})=0$
$\Rightarrow\cos (\text{A}+4\text{B})= \cos90^\circ$
$\Rightarrow\text {A}+4\text{B}=90^ \circ\dots(2)$
Substract (1) from (2), we get
$\Rightarrow2\text{B} =30^\circ$
$\Rightarrow\text{B}= \frac{30^\circ} {2}=15^\circ$
Put B = 15° in (1) we get -
$\Rightarrow\text {A}+2\times15=60^ \circ$
$\Rightarrow\text {A}+30^\circ=60^\circ $
$\Rightarrow\text{A} =60^\circ-30^\circ $
$\Rightarrow\text{A} =30^\circ$
Thus, A = 30° and B = 15°
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