Question

The angle between the lines $x\cos {30}^o+y\sin {30}^o=3$
$x\cos {60}^o+y\sin {60}^o=5$ is

• A. ${90}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. None of these

Answer 1

The correct option is B ${30}^o$

$xcos{30}^{\circ }+ysin{30}^{\circ }=3$

$y=-xcot{30}^{\circ }+3cosec{30}^{\circ }$

$m_1=-cot{30}^{\circ }=-\sqrt{3}$

$xcos{60}^{\circ }+ysin{60}^{\circ }=5$

$y=-xcot{60}^{\circ }+5cosec{60}^{\circ }$

$m_2=-cot{60}^{\circ }=-1/\sqrt{3}$

$tan\theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{-3+1/\sqrt{3}}{1+(-1/\sqrt{3})(-1/\sqrt{3})}\right|$

$=\left|\frac{-3+1}{\sqrt{3}(1+1)}\right|=\left|\frac{1}{\sqrt{3}}\right|$

$\theta ={30}^{\circ }$