If $sec\ 4A = cosec\ (A - 20^o)$, where $4A$ is an acute angle, find the value of $A$.

Given:

\( \sec 4 A=\operatorname{cosec}\left(A-20^{\circ}\right) \), where \( 4 A \) is an acute angle.

To do:

We have to find the value of \( A \).

Solution:  

We know that,

$\operatorname{cosec} (90^{\circ}- \theta) = sec\ \theta$

Therefore,

$\sec 4 A=\operatorname{cosec}\left(A-20^{\circ}\right)$

$\operatorname{cosec} (90^{\circ}- 4A)=\operatorname{cosec}\left(A-20^{\circ}\right)$

Comparing on both sides, we get,

$90^{\circ}-4A =A-20^{\circ}$

$4A+A=90^{\circ}+20^{\circ}$

$5A=110^{\circ}$

$A=\frac{110^{\circ}}{5}$

$A=22^{\circ}$

The value of \( A \) is $22^{\circ}$.    

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Updated on: 10-Oct-2022

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