Question
If $5\text{A}=\text{cosec }(\text{A}-36^\circ)$ and $5\text{A}$ is an acute angle, show that $\text{A}=21^\circ.$
✓
Answer
Given: $\sec\text{5A}=\text{cosec}(\text{A}-36^\circ)$$\Rightarrow\text{cosec}(90^\circ-\text{5A})=\text{cosec}(\text{A}-36^\circ)$ $\big[\because\text{cosec}(90^\circ-\theta)=\sec\theta\big]$
$\Rightarrow90^\circ-\text{5A}=\text{A}-36^\circ$
$\Rightarrow\text{6A}=90^\circ+36^\circ$
$\Rightarrow\text{6A}=126^\circ$
$\Rightarrow\text{A}=21^\circ$
$\Rightarrow90^\circ-\text{5A}=\text{A}-36^\circ$
$\Rightarrow\text{6A}=90^\circ+36^\circ$
$\Rightarrow\text{6A}=126^\circ$
$\Rightarrow\text{A}=21^\circ$
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