If $ \sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right) $, where $ 2 A $ is an acute angle, find the value of $ A $.

Given:

\( \sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right) \), where \( 2 A \) 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,

$\sec 2 A=\operatorname{cosec}\left(A-42^{\circ}\right)$

$\operatorname{cosec} (90^{\circ}- 2A)=\operatorname{cosec}\left(A-42^{\circ}\right)$

Comparing on both sides, we get,

$90^{\circ}-2A =A-42^{\circ}$

$2A+A=90^{\circ}+42^{\circ}$

$3A=132^{\circ}$

$A=\frac{132^{\circ}}{3}$

$A=44^{\circ}$

The value of \( A \) is $44^{\circ}$.   

Tutorialspoint

Updated on: 10-Oct-2022

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