Question

If $0^0<\theta <{90}^0$ find $\theta$ $2\cos \theta +2\sqrt{2}=3\sec \theta$

• A. ${45}^0$
• B. ${65}^0$
• C. ${60}^0$
• D. None of these

Answer 1

Answer: (A)

${45}^0$

$2\cos \theta +2\sqrt{2}=3\sec \theta =\frac{3}{\cos \theta }$
or $2{\cos }^2\theta +2\sqrt{2}\cos \theta -3=0$
$\therefore \cos \theta =\frac{-2\sqrt{2}+\sqrt{{(2-\sqrt{2})}^2+4\times 2\times 3}}{2\times 2}$
$=\frac{-2\sqrt{2}\pm 4\sqrt{2}}{4}=\frac{2\sqrt{2}}{4}$ or $\frac{-6\sqrt{2}}{4}$
$=\frac{1}{\sqrt{2}}$ or $-\frac{3\sqrt{2}}{4}$ or $\theta ={45}^0$