Question

One vertex of an equilateral triangle is origin and $OX$ (where $X$ lies on the $x-$axis) is one side with length $a$. The remaining two vertices, in $(r,\theta )$ form, are

• A. $(a,0),(a,\frac{\pi }{3})$
• B. $(a,\frac{\pi }{2}),\left(a\frac{\pi }{4}\right)$
• C. $(a,0),(a,\frac{\pi }{2})$
• D. $(0,0),(a,\frac{\pi }{2})$

Answer 1

The correct option is A $(a,0),(a,\frac{\pi }{3})$

Given one vertex of the triangle is $(0,0)$
One vertex lies at $x-$axis at a distance $a$ from origin. So, the other vertex is $A(a,0)$.
Since, the triangle is equilateral,
$\Rightarrow OA=OB=AB=a$
So, let the third vertex be $B(a,\theta )$.
Now, $AB=\sqrt{a^2+a^2-2a^2\cos \theta }$
$\Rightarrow a^2=2a^2-2a^2\cos \theta$
$\Rightarrow a^2(2\cos \theta -1)=0$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$
Hence, the other two vertices are $(a,0)$ and $(a,\frac{\pi }{3})$.