Question

In the diagram given, equation of $AB$ is $x=\sqrt{3}y+1=0$ and equation of $AC$ is $x-y-2=0$
(i) Write down the angles that the lines $AC$ and $AB$ make with the positive direction of x-axis
(ii) Find $\angle BAC$

Answer 1

Slope of AB: $\sqrt{3}y-x+1=0$

$\Rightarrow -\frac{1}{\sqrt{3}}=\tan \theta$

$\Rightarrow {\theta }_1={30}^o$ (where $\theta$ is angle which AB makes with positive x-axis). Similarly slope of AC$=\frac{-1}{-1}=1$

$\Rightarrow \tan \theta =1\Rightarrow {\theta }_2={45}^o$

$\Rightarrow \angle ABC={30}^o,\angle ACB=180-45={135}^o$

$\Rightarrow \angle BAC=180-30-135={15}^o$.