Question

The angle between the lines $x\cos 30°+y\sin 30°=3$and $x\cos 60°+y\sin 60°=5$is

• A. $90°$
• B. $30°$
• C. $60°$
• D. None of these

Answer 1

The correct option is B

$30°$

Explanation for correct answer:

Finding the angle between the lines:

Given, the equation as $x\cos 30°+y\sin 30°=3$ and $x\cos 60°+y\sin 60°=5$ are two lines.

The equation can be written as

$\begin{array}{rcl}x\left(\frac{\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)& =& 3\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{30}\mathbf{°}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{30}\mathbf{°}\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\right]\\ \frac{\sqrt{3}}{2}x+\frac{1}{2}y& =& 3\to \left(i\right)\end{array}$

and

$\begin{array}{rcl}x\left(\frac{1}{2}\right)+y\left(\frac{\sqrt{3}}{2}\right)& =& 5\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{60}\mathbf{°}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{60}\mathbf{°}\mathbf{=}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\right]\\ \frac{1}{2}x+\frac{\sqrt{3}}{2}y& =& 5\to \left(ii\right)\end{array}$

Now find the slope of the equation as $m=\frac{-a}{b}$

Finding the slope of the equation $\left(i\right)$

$\begin{array}{rcl}{m}_1& =& \frac{-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}\\ m_1& =& -\sqrt{3}\end{array}$

And finding the slope of the equation $\left(ii\right)$

$\begin{array}{rcl}{m}_2& =& \frac{-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\\ m_2& =& -\frac{1}{2}\times \frac{2}{\sqrt{3}}\\ & =& -\frac{1}{\sqrt{3}}\end{array}$

We know that,

$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{=}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\begin{array}{l}\tan \theta =\left|\frac{-\sqrt{3}-\left(-\frac{1}{\sqrt{3}}\right)}{1+\left(-\sqrt{3}\right)\left(-\frac{1}{\sqrt{3}}\right)}\right|\\ =\left|\frac{\frac{-{\left(\sqrt{3}\right)}^2+1}{\sqrt{3}}}{1+1}\right|\\ =\left|\frac{-3+1}{2\sqrt{3}}\right|\\ =\left|\frac{-2}{2\sqrt{3}}\right|\\ =\left|\frac{-1}{\sqrt{3}}\right|\\ =\frac{1}{\sqrt{3}}\end{array}$

$\begin{array}{rcl}\tan \theta & =& \frac{1}{\sqrt{3}}\\ \theta & =& {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ \theta & =& 30°\end{array}$

Therefore, the angle between the two lines is $30°$.

Hence, option (B) is the correct answer