Find the acute angle 0 such that 5 2 0 + 3 = 9 0. - Vidyadip

Question

Find the acute angle $0$ such that $5 \tan 2 0 + 3 = 9 \sec 0.$

✓

Answer

$5 \tan ^2 \theta+3=9 \sec \theta$
$\therefore 5\left(\sec ^2 \theta-1\right)+3=9 \sec \theta$
$\therefore 5 \sec ^2 \theta-5+3=9 \sec \theta$
$\therefore 5 \sec ^2 \theta-9 \sec \theta-2=0$
$\therefore 5 \sec ^2 \theta-10 \sec \theta+\sec \theta-2=0$
$\therefore 5 \sec \theta(\sec \theta-2)+1(\sec \theta-2)=0$
$\therefore(\sec \theta-2)(5 \sec \theta+1)=0$
$\therefore \sec \theta-2=0 \text { or } 5 \sec \theta+1=0$
$\therefore \sec \theta=2 \text { or sec } \theta=-1 / 5$
$\text { Since sec } \theta \geq 1 \text { or sec } \theta \leq-1,$
$\sec \theta=2$
$\therefore \theta=60^{\circ} \ldots\left[\because \sec 60^{\circ}=2\right]$

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