Question

The perimeter of the triangle formed by the points $(0,0),(3,\frac{\pi }{3})$ and $(3,\frac{2\pi }{3})$ is?

• A. $3$
• B. $6$
• C. $9$
• D. $18$

Answer 1

The correct option is C $9$

The perimeter of triangle $△ABC$ is the summation of the length of sides AB, BC, CA.

The discordance between two points is:

$d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos ({\theta }_1-{\theta }_2)}$

length of AB where A(0,0) and B(3, $\frac{\pi }{3}$)

$r_1=0,r_2=3,{\theta }_2=\frac{2\pi }{3},{\theta }_1=\frac{\pi }{3}AB=\sqrt{0^2+3^2-2\times 3\times 0\cos (\frac{2\pi }{3}-\frac{\pi }{3})}AB=\sqrt{0+9-0}$

$AB=3$units

length of BC where B(3,\frac{\pi}{3}) and C(3, $\frac{2\pi }{3}$)

$r_1=3,r_2=3,{\theta }_1=\frac{\pi }{3},{\theta }_2=\frac{2\pi }{3}BC=\sqrt{3^2+3^2-2\times 3\times 3\cos (\frac{\pi }{3}-\frac{2\pi }{3})}BC=\sqrt{9+9-18\cos \left(\frac{-\pi }{3}\right)}$

$BC=3$units

length of CA where C(3, $\frac{2\pi }{3}$) and A(0, 0)

$r_1=3,r_2=0,{\theta }_1=\frac{\pi }{3},{\theta }_2=\frac{2\pi }{3}CA=\sqrt{3^2+0^2-2\times 3\times 0\cos (\frac{\pi }{3}-\frac{2\pi }{3})}CA=\sqrt{9+0-0}$

$CA=3$units

Perimeter $=AB+BC+CA=3+3+3+=9\text{units}$