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In an acute angled triangle $ABC$, if $\tan \left(A+B-C \right)=1$ and  $\sec \left(B+C-A \right)=2$, Find angle $A$.

  1. $60^\circ$
  2. $45^\circ$
  3. $30^\circ$
  4. $90^\circ$
in Quantitative Aptitude 9.2k points 27 445 804
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1 Answer

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Given: $\tan (A+B-C)=1$

$\implies \tan(A+B-C)=\tan 45^\circ$

$\implies (A+B-C)=45^\circ$ …...(i)

In same way $\sec(B+C-A)=2$

$\implies \sec(B+C-A)=\sec 60^\circ$

$\implies (B+C-A)=60^\circ$……..(ii)

from equation (i) & (ii)

                         $2B=105^\circ\implies B=52.5^\circ$

For any triangle

$\because \angle A+\angle B+\angle C=180^\circ$

$\implies A+B=180^\circ-C$…..(iii)

from equation (i)&(iii),

$\implies 180^\circ-C-C=45^\circ$

$\implies 2C=135^\circ \implies 67.5^\circ$

now form equation (iii) we get:

$\angle A+52.5^\circ=180^\circ-67.5^\circ$

$\angle A=60^\circ$

Option $A$ is correct here.
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